Introductory Circuit Theory 3030319849, 9783030319847, 9783030319854

Table of contents :
Preface......Page 5
Contents......Page 7
Abbreviations......Page 10
1 Basic Concepts......Page 11
1.1 Ohm's Law......Page 13
1.2 Resistors Connected in Series......Page 14
1.3 Resistors Connected in Parallel......Page 17
1.4 Resistors Connected in Series and Parallel......Page 19
1.5 Kirchhoff's Voltage and Current Laws......Page 21
1.7 Summary......Page 23
Exercises......Page 24
2 DC Circuits......Page 28
2.1 Mesh Analysis......Page 29
2.2 Nodal Analysis......Page 36
2.3 Examples......Page 38
2.3.1 Linearity Property of Circuits......Page 46
2.3.2 Analysis of a Circuit with a Controlled Voltage Source......Page 49
2.3.3 Analysis of a Circuit with a Controlled Current Source......Page 53
Circuit in Fig.2.22......Page 55
Circuit in Fig.2.23......Page 57
Circuit in Fig.2.24......Page 58
Circuit in Fig.2.25......Page 59
Circuit in Fig.2.26......Page 60
Circuit in Fig.2.27......Page 61
Circuit in Fig.2.28......Page 62
Circuit in Fig.2.29......Page 63
2.3.4 Y- and -Y Transformations......Page 64
Duality......Page 65
Thévenin's Theorem......Page 68
Norton's Theorem......Page 70
2.4.2 Maximum Power Transfer Theorem......Page 74
2.5.1 Strain Gauge Measurement......Page 76
2.6 Summary......Page 77
Exercises......Page 79
3.1 Sinusoids......Page 85
3.1.2 The Complex Sinusoids......Page 87
3.2.1 Time- and Frequency-Domain Representations of Circuit Elements......Page 88
3.2.2 Time-Domain Analysis of a Series RC Circuit......Page 90
3.2.3 Frequency-Domain Analysis of a RC Circuit......Page 92
3.2.4 Impedances Connected in Series......Page 93
3.2.5 Impedances Connected in Parallel......Page 99
3.2.6 Impedances Connected in Series and Parallel......Page 101
The Source......Page 102
The Impedances......Page 103
Verifying the Solutions......Page 104
Circuit 3.2......Page 106
Circuit 3.3......Page 108
Circuit 3.4......Page 110
Circuit 3.5......Page 112
3.2.8 Linearity Property of Circuits......Page 113
Circuit 3.6......Page 115
Circuit 3.7......Page 119
Circuit in Fig.3.21......Page 121
Circuit in Fig.3.22......Page 123
Circuit in Fig.3.23......Page 124
Circuit in Fig.3.24......Page 125
Circuit in Fig.3.25......Page 126
Circuit in Fig.3.26......Page 127
Circuit in Fig.3.27......Page 129
3.3.1 Thévenin's Theorem......Page 130
3.3.2 Norton's Theorem......Page 138
3.3.3 Maximum Average Power Transfer Theorem......Page 139
3.3.4 Source Transformation......Page 141
RL Filter......Page 143
RC Filter......Page 144
Exercises......Page 145
4 Steady-State Power......Page 157
4.1 Energy in Reactive Elements with Sinusoidal Sources......Page 158
4.2 Power Relations in a Circuit......Page 159
4.3 Power-Factor Correction......Page 167
4.5 Summary......Page 181
Exercises......Page 182
5.1 Mutual Inductance......Page 184
5.2 Stored Energy......Page 185
5.3 Examples......Page 188
Ideal Transformers......Page 205
5.5 Summary......Page 206
Exercises......Page 207
6.1 Three-Phase Voltages......Page 209
6.2 The Three-Phase Balanced Y - Y Circuit......Page 211
6.3 The Three-Phase Balanced Y- Circuit......Page 213
6.4 The Three-Phase Balanced -Y Circuit......Page 216
6.5 The Three-Phase Balanced - Circuit......Page 217
6.7 Summary......Page 218
Exercises......Page 219
7.1 Impedance Parameters......Page 221
7.2 Admittance Parameters......Page 227
7.3 Hybrid Parameters......Page 230
7.4 Transmission Parameters......Page 231
7.5.1 Analysis of a π Circuit......Page 232
7.5.2 Analysis of a Common-Emitter Transistor Amplifier......Page 234
7.5.3 Analysis of a Bridge Circuit......Page 236
7.5.4 Ladder Circuit......Page 237
7.6.1 Digital-to-Analog Converter: The R-2R Ladder Circuit......Page 239
The Number System......Page 240
R-2R Ladder Circuit......Page 241
Exercises......Page 242
8 Transform Analysis and Transient Response......Page 244
8.1 Fourier Series......Page 245
8.1.1 Fourier Series of a Rectified Sine Wave......Page 247
8.2 Fourier Transform......Page 251
8.2.1 The Transfer Function and the Frequency Response......Page 253
8.3 Transient Response......Page 254
8.3.1 The Unit-Impulse and Unit-Step Signals......Page 255
The Unit-Impulse as the Derivative of the Unit-Step......Page 256
8.4 Laplace Transform......Page 257
Linearity......Page 258
8.4.3 Integration......Page 259
8.4.6 Circuit Analysis in the Frequency-Domain......Page 260
Analysis of Circuits with Arbitrary Initial Conditions......Page 269
8.5 Application......Page 274
8.6 Summary......Page 276
Exercises......Page 278
A Matrices......Page 280
A.1 Determinants......Page 282
B Complex Numbers......Page 285
Answers to Selected Exercises......Page 289
Chapter 3......Page 290
Chapter 4......Page 292
Chapter 6......Page 293
Chapter 8......Page 294
Bibliography......Page 295
Index......Page 296

Citation preview

D. Sundararajan

Introductory Circuit Theory

Introductory Circuit Theory

D. Sundararajan

Introductory Circuit Theory

D. Sundararajan Formerly at Concordia University Montreal, QC, Canada

Additional material to this book can be downloaded from https://Springer.com. ISBN 978-3-030-31984-7 ISBN 978-3-030-31985-4 (eBook) https://doi.org/10.1007/978-3-030-31985-4 © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Circuit theory is fundamental to the study of important engineering applications such as power systems and communication. As such, it forms part of the first course to engineering and science students in several disciplines. This book is written for engineering, computer science and physics students, and engineers and scientists. Both the theory and practice by programming and in laboratory are two important aspects of the learning process of circuit theory. The objective in writing this book is to present the fundamentals of circuit theory systematically in a clear and concise textbook and provide the required programming part online. Hopefully, this approach is expected to improve the readability and understandability of the theory without clutter. Plenty of examples, figures, tables, programs, analogy, and physical explanations make it easy for the reader to get a good grounding in the basics of circuit theory and some of its applications. The learning of the circuit theory requires calculus, linear algebra, transform analysis, programming, and laboratory practice. These tools are also required for other courses and in professional career later. However, these topics are difficult for new students. But, through the methods suggested, students can learn well this all-important subject with sufficient practice. Further, the learning process for this subject will certainly help the students in their ensuing study of the other subjects. Learning of the circuit theory consists of four aspects: (1) systematic presentation of the mathematical methods in the text; (2) verification of the analytically obtained results by coding; (3) verification of the results by simulation; and (4) verification of the results using actual components in laboratory experiments. The last component gives physical appreciation of the circuit elements and their values in practice, the voltages, currents, frequencies, measuring instruments, and all other practical aspects involved. Students must take a coordinated laboratory course. Both coding and simulation of the analysis of circuits are presented in the online programming part. Each student can practice using these four methods as much as it is required for a good understanding of the subject. This book is intended to be a textbook for the first course in circuit theory to the new undergraduate students in several disciplines of engineering and science, which includes the primary disciplines of electrical and electronics engineering. For engineering professionals, this book will be useful for selfstudy. In addition, this book will be a reference for anyone, student or professional, specializing in practical applications of circuit theory. The prerequisite for reading this book is a good knowledge of physics and calculus at the high school level. As mentioned already, programming is an important component in learning and practicing circuit theory, as well as other subjects. While several software packages are available, it is better to use a popular general-purpose software package that the students are likely to use in their other courses and ensuing professional carrier in several areas of engineering and science. Therefore, learning of only one package is required. The programming part is presented using the popular, user-friendly R software package. While the use and widely used, both in universities and industries, MATLAB of a software package is inevitable in most applications, it is better to use the software in addition to v

vi

Preface

self-developed programs. The effective use of a software package or to develop own programs requires a good grounding in the basic principles of the circuit theory. Answers to selected exercises marked ∗ are given at the end of the book. A Solutions Manual and slides are available for instructors at the website of the book. I assume the responsibility for all the errors in this book and would very much appreciate receiving readers’ suggestions and pointing out any errors (email:[email protected]). I am grateful to my Editor and the rest of the team at Springer for their help and encouragement in completing this project. I thank my family for their support during this endeavor. D. Sundararajan

Contents

1

Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Resistors Connected in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Resistors Connected in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Resistors Connected in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Kirchhoff’s Voltage and Current Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 3 4 7 9 11 13 13

2

DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Mesh Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Nodal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Linearity Property of Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Analysis of a Circuit with a Controlled Voltage Source . . . . . . . . . . . . . . . . . . . 2.3.3 Analysis of a Circuit with a Controlled Current Source . . . . . . . . . . . . . . . . . . . 2.3.4 Y −  and  − Y Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Circuit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Thévenin’s Theorem and Norton’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Maximum Power Transfer Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Strain Gauge Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19 20 27 29 37 40 44 55 59 59 65 67 67 68

3

AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.1 Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.1.1 The Rectangular Form of Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.1.2 The Complex Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.2 AC Circuit Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.2.1 Time- and Frequency-Domain Representations of Circuit Elements . . . . . . . . . 80 3.2.2 Time-Domain Analysis of a Series RC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 3.2.3 Frequency-Domain Analysis of a RC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 3.2.4 Impedances Connected in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 3.2.5 Impedances Connected in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.2.6 Impedances Connected in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 3.2.7 Analysis of Typical Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 3.2.8 Linearity Property of Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 vii

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3.3

Circuit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Thévenin’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Norton’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Maximum Average Power Transfer Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Source Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

122 122 130 131 133 135 135 137

4

Steady-State Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Energy in Reactive Elements with Sinusoidal Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Power Relations in a Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Power-Factor Correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149 150 151 159 173 173

5

Magnetically Coupled Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Stored Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

177 177 178 181 198 198 199

6

Three-Phase Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Three-Phase Voltages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 The Instantaneous Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Three-Phase Balanced Y − Y Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Three-Phase Balanced Y −  Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Three-Phase Balanced  − Y Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The Three-Phase Balanced  −  Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

203 203 205 205 207 210 211 212 212

7

Two-Port Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Impedance Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Admittance Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Hybrid Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Transmission Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Analysis of a π Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.2 Analysis of a Common-Emitter Transistor Amplifier . . . . . . . . . . . . . . . . . . . . . 7.5.3 Analysis of a Bridge Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.4 Ladder Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Digital-to-Analog Converter: The R − 2R Ladder Circuit . . . . . . . . . . . . . . . . . 7.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

215 215 221 224 225 226 226 228 230 231 233 233 236

3.4 3.5

Contents

8

Transform Analysis and Transient Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Fourier Series of a Rectified Sine Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Gibbs Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 The Transfer Function and the Frequency Response . . . . . . . . . . . . . . . . . . . . . . 8.3 Transient Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 The Unit-Impulse and Unit-Step Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Properties of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Time-Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.4 Initial Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.5 Final Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.6 Circuit Analysis in the Frequency-Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

239 240 242 246 246 248 249 250 252 253 254 254 255 255 255 269 271

A Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 A.1 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 B Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

Abbreviations

AC DC FS FT Im KCL KVL LTI pf RC Re RL RLC RMS

Alternating current, sinusoidally varying current or voltage Direct current, sinusoid with frequency zero, constant current or voltage Fourier series Fourier transform Imaginary part of Kirchhoff’s Current Law Kirchhoff’s Voltage Law Linear time-invariant Power factor Circuit comprising a resistor and a capacitor in series Real part of Circuit comprising a resistor and an inductor in series Circuit comprising a resistor, a capacitor, and an inductor in series Root-mean-square value

xi

Basic Concepts

1

Electrical and electronic engineering is indispensable in all applications of science and engineering, either in home, office, or industry. In applications, signals occur in different forms, such as temperature, pressure, audio, video, medical, and optical. All the signals are converted to electrical signals using appropriate transducers, which convert signals from other forms to electrical, for easier and efficient analysis, control, generation, and transmission. For example, a microphone converts sound waves into an electrical signal. A turbogenerator, a steam turbine coupled to an electric generator, produces electrical power. Windmills serve a similar purpose. Once the signals are available in electrical form, there are two major types of activities in electrical and electronic engineering those are indispensable for all applications, power systems and communication. For these and other types of applications, circuit theory is of fundamental importance. That is why, this subject is learnt by students from several disciplines, in addition to the primary disciplines of electrical and electronic engineering. By analogy, the basic principles of circuit theory are applicable to other systems, such as mechanical, optical, and acoustical. The effects of electrical power are clearly visible in the bulb emitting light, a water heater making the water hot and a fan creating a current of air. However, the flow of electric current is invisible and this aspect makes it difficult to visualize it. Fortunately, it is similar to the flow of water in a pipe. This analogy enables us to visualize the current flow in an electrical system. The analogy is similar to the operation of a computer providing a good analogy to the working of the brain. Water head determines the flow of water through a hose and pinching it increases the resistance to the flow. Parallel pipes decrease the resistance to the flow and pipes in series increase the resistance. Water towers provide the pressure to supply water to our homes. When the tank is full, the flow is more compared with when it is near empty. Water pressure is the analog to electrical voltage and water flow is the analog to current. Figure 1.1a shows water flowing from a tap, with the tap partly open. Figure 1.1b shows water flowing from the tap, with the tap fully open. In this case, the resistance to the flow of water is relatively reduced. For the same opening of the tap, the flow can be increased by increasing the water pressure from the overhead tank. Electricity is the movement of charged particles, such as electrons. A circuit is a closed loop, which allows the flow of charges from one place to another. Components in the circuit control the flow and use it to do work. Charge is similar to the amount of water. Voltage is similar to water pressure. Current is flow of charge. More water in the tank results in increasing the flow at the exit. Flash light gets dimmer as the batteries run down. Less pressure results in reduced water flow. The measure of water flow is cubic meter/second. The measure of current flow is coulomb/second. © Springer Nature Switzerland AG 2020 D. Sundararajan, Introductory Circuit Theory, https://doi.org/10.1007/978-3-030-31985-4_1

1

2

1 Basic Concepts

Fig. 1.1 (a) Water flow with the tap partly open; (b) more flow with the tap fully open

(a)

(b)

Fig. 1.2 Batteries in series

Bulb V

V Dim V Bulb

(a)

2V Bright

V

(b) An electrical system is similar to a water supply system. Water flows through the pipes of a water system. Electricity flows through the wires of an electrical system. Electrical pressure is measured in volts. Electrical current flow is measured in amperes. A wire serves the purpose of a pipe. A battery serves the purpose of a pump or a reservoir. A resistor serves the purpose of a narrow constriction in a pipe. Current flow is similar to water flow. Current is the number of electrons moving past a point in the circuit per second. Voltage is the pressure difference between two points, due to which the current flows. A larger conductor offers less resistance and enables a larger current flow in the circuit with the same voltage applied. With the resistance remaining the same, applying a higher voltage results in a larger current flow, as shown in Fig. 1.2. In Fig. 1.2a, V volts of electrical pressure is applied to the bulb and the resulting light is dim. In (b), two batteries connected in series produce 2 V volts of pressure resulting in the bulb becoming bright. With a higher voltage applied to the circuit, more current flows making the bulb emit more light. A water supply system has water tank, pumps, pipes connected in various configurations, and valves. Similarly, an electric circuit has voltage and current sources, wires and circuit elements, connected in various configurations. No analogy can be exact. Here is another one. Cars cross tollgate in the highway. If there is only one gate, certain number of cars can cross the gate per unit time. If the cars have to cross two gates, one after another, then the rate of cars passing through the gates becomes one-half. If there are two parallel gates, double the number of cars can pass through. If there are only few cars, then one gate may be sufficient. But, a large number of cars forces opening of more gates resulting in a higher rate of flow. In electrical systems, this is called Ohm’s law. A higher voltage produces a larger current in the circuit. Processing time at the gate has to be reduced by opening more gates. Similarly, a higher voltage has to be applied for more current and more light from the bulb. Similarly, a larger conductor offers less resistance and enables a larger current flow in the circuit with the same voltage applied.

1.1 Ohm’s Law

3

V = 4V, R = 2Ω, G = 0.5 , I = 42 = 4(0.5) = 2A I = GV +

Fig. 1.3 Resistor R in ohm or conductance G in mho

V = RI

1 R= G

A voltage is developed by creating a separation of positive and negative charges. One coulomb of charge is the total charge of 6.242 × 1018 electrons. To develop 1 volt (V) between two points, 1 coulomb of charge has to be moved by applying 1 joule of energy. This is similar to pumping water to a higher level. Since potential energy is due to position, the word potential is also used to refer voltage levels. When a connection is made between two points with a potential difference, current flows. If 1 coulomb of charge passes through a point per second, then the flow of charge or current is said to be 1 ampere (A). Typical voltage sources are batteries, solar cells, and generators. In a practical electrical circuit, the elements are physical devices. For the purpose of analysis, any system can be represented by mathematical models with acceptable tolerances. Therefore, the elements of an electric circuit are represented by mathematical definitions. The opposition to the flow of current is called as the resistance. The resistance is 1  if 1 ampere current flows through it, when 1 V is applied across it. The resistor, a device used to control the flow of current, is denoted by the symbol R. The reciprocal of resistance is conductance, denoted by G. That is, R=

1 G

and

G=

1 . R

The unit of measurement of resistance is , called ohm. The unit of measurement of conductance is , called mho. Figure 1.3 shows the input–output relationship of a resistor of value R. With 4 V applied across the resistor of value 2 , the current I through it is 4/2 = 2 A. The conductance is 1/2 = 0.5. Then, I = 4 × 0.5 = 2 A, as found earlier. The resistance is similar to friction. If a surface is rough, more force is required to move an object on it. The more the roughness of the surface, the more is the heat generated. The higher the current flowing through the resistor, the higher is the voltage across it. Both the frictional devices and resistors dissipate energy, when a force is applied.

1.1

Ohm’s Law

The voltage across a resistor, V , is the current, I , flowing through it times its value R, known as Ohm’s law. That is, V V and V = I R. I= , R= R I It is assumed that R is a constant to a required accuracy, since R varies with temperature, pressure, etc. This law, as is the case with most of the other laws, is applicable to both the DC and AC circuits. By convention, the current enters the positive terminal and leaves the negative terminal. That is, current flows from a higher voltage to a lower voltage. Therefore, the current direction determines the polarity of the voltage at the terminals of the resistor. If R is zero, it is called a short-circuit and the voltage across it is zero. If R is ∞, it is called an open-circuit and the current through it is zero. Resistors are essential to control the flow of current and are commonly used in such applications as controlling the speed of a fan, audio volume control in amplifiers, and emission of light by a bulb.

4

1 Basic Concepts

The power dissipated by a resistor, the rate of energy dissipation, is P = VI =

V2 = I 2 R. R

The unit of measurement is joules per second or watts (W). Note that the expressions for the power are quadratic. That is, they are nonlinear. The power varies inversely as the resistance with a voltage source. The power varies proportionally as the resistance with a current source. The power consumed in a circuit is the sum of the powers consumed by all the constituent resistors, found using either of the defining expressions. Alternatively, when there is only one source, the equivalent resistance of the whole circuit can be computed. Then, the power dissipated in this resistor is the power consumed by the whole circuit. When more than one source is present, the power consumed by the circuit is the sum of the powers consumed by each resistor. Example 1.1 Determine the resistance of a bulb, when the current through it is 0.2 A and the voltage across it is 220 V. Find the power consumed by the bulb. Solution R=

220 = 1100 0.2

P = 220 × 0.2 =

2202 = 0.22 1100 = 44W. 1100 

Example 1.2 Determine the resistance of a smoothing iron, when the current through it is 10 A and the voltage across it is 110 V. Find the power consumed by the iron. Solution R= P = 110 × 10 =

110 = 11 10 1102 = 102 11 = 1100W. 11 

1.2

Resistors Connected in Series

In practice, the desired resistor is often not available and we have to use a combination of more than one resistor to make an equivalent one. Resistors can be connected in series and/or parallel configurations. Resistors have two terminals and, therefore, they come under the class of two-terminal devices or elements. In a series connection, one, and only one, terminal of a resistor is connected to adjoining resistors. Figure 1.4a shows two resistors connected in series, called a series circuit. A circuit is an interconnection of elements. The determination of current and voltages at all parts of the circuit is the essence of circuit analysis. When resistors are connected in series, the voltage across them increases, with the same current flowing through them. It is similar to connecting hoses to make

1.2 Resistors Connected in Series

5

VR1

VR1 +

R1 V +

R2 VR2

VR1 = V VR2 = V

3Ω

10V +

−

− I +

+ I − 3Ω VR2 +

+

2 = 2V VR1 = 5 2+3

2 VR2 = 10 3+2 = 4V

(a)

2Ω

5V

2Ω VR2 −

3 = 6V VR1 = 10 3+2

R1 R1 +R2 R2 R1 +R2

VR1

3 VR2 = 5 2+3 = 3V

(b)

(c)

Fig. 1.4 Series circuit with a voltage source

a longer hose. The combined resistance is the sum of all the resistances. That is, with N number of resistors connected in series, the equivalent resistance Req of the series circuit is Req = R1 + R2 + · · · + RN . The value of Req will be larger than the largest resistor in the series connection. The same current I pass through all the resistors. Therefore, the voltage V across the series connection is V = I R1 + I R2 + · · · + I RN = I Req . The total resistance remains unchanged, irrespective of the order in which they are connected. Obviously, if all of them have the same value, then Req = N R. The source voltage applied across them gets divided in proportion to their individual values. The current through the series circuit is I=

V Req

and the voltage across any resistor Rn is VRn =

V Rn . Req

With just two resistors, R1 and R2 , and V , the voltages across the series connection are VR1 = V

R1 R1 + R2

and

VR2 = V

R2 . R1 + R2

Consider the circuit shown in Fig. 1.4b. The circuit is energized by a voltage source of 10 V. An ideal voltage source maintains a constant voltage at its terminals, irrespective of the current drawn from it. A voltage source is a constraint, clamping the voltage at a certain point in the circuit. The voltage drops across the resistors are VR1 =

V 10 3 = 6V R1 = R1 + R2 3+2

and

VR2 =

V 10 2 = 4 V. R2 = R1 + R2 3+2

6

1 Basic Concepts

VR1

VR1 +

R1 IA

R2 VR2

3Ω

2A

−

− + 2Ω VR2 −

VR1 2Ω

1A

+ − 3Ω VR2 +

VR1 = IR1

VR1 = 2 × 3 = 6V

VR1 = 1 × 2 = 2V

VR2 = IR2

VR2 = 2 × 2 = 4V

VR2 = 1 × 3 = 3V

(a)

(b)

(c)

Fig. 1.5 Series circuit with a current source

The voltage drops add up to the source voltage 10 = 6 + 4. The current through the circuit is 2 A. The power consumed by the circuit is

10 3+2

=

P = 10 × 2 = 20 W. Consider the circuit shown in Fig. 1.4c. The circuit is energized by a voltage source of 5 V. As the polarity of the source voltage is reversed, the polarity of the voltage drops across the resistors is also reversed. The voltage drops across the resistors are VR1 =

V 5 R1 = 2 = 2V R1 + R2 3+2

and

VR2 =

V 5 R2 = 3 = 3V R1 + R2 3+2

5 = The voltage drops add up to the source voltage 5 = 2 + 3. The current through the circuit is − 3+2 −1 A. The power consumed by the circuit is

P = −5 × −1 = 5 W Consider the circuit shown in Fig. 1.5a. The circuit is energized by a current source of I A. An ideal current source maintains a constant current at its terminals, irrespective of the voltage across its terminals. A current source is a constraint, clamping the current at a certain point in the circuit. As the same current flows through the two resistors, the respective voltage drops are VR1 = I R1

and

VR2 = I R2 .

Consider the circuit shown in Fig. 1.5b. The circuit is energized by a current source of I = 2 A. The respective voltage drops are VR1 = I R1 = 2 × 3 = 6 V

and

VR2 = I R2 = 2 × 2 = 4 V.

The power consumed by the circuit is P = 10 × 2 = 20 W.

1.3 Resistors Connected in Parallel

7

Consider the circuit shown in Fig. 1.5c. The circuit is energized by a current source of I = 1 A. As the direction of current flow is reversed, the polarities of the voltage drops are also reversed. The respective voltage drops are VR1 = I R1 = 1 × 2 = 2 V

VR2 = I R2 = 1 × 3 = 3 V.

and

The power consumed by the circuit is P = −5 × −1 = 5 W.

1.3

Resistors Connected in Parallel

In a parallel connection of elements, all of them are connected such that they have two points in common. Figure 1.6 shows two resistors connected in parallel, called a parallel circuit. When resistors are connected in parallel, the voltage across all of them remains the same, with different currents flowing through them. It is similar to connecting hoses to make a wider hose. The length remains the same, but the flowing capacity increases. The combined resistance is the reciprocal of the sum of all the conductances. That is, with N number of resistors connected in parallel, the equivalent resistance Req of the parallel circuit is Geq = G1 + G2 + · · · + GN

and

Req =

1 , Geq

where Gn = 1/Rn . The value of Req will be smaller than the smallest resistor in the parallel connection, since the total current is more. The same voltage V is applied across all the resistors. Therefore, the total current I flowing through the parallel connection is I = V G1 + V G2 + · · · + V GN = V Geq . The total resistance remains unchanged, irrespective of the order in which they are connected. Obviously, if all of them have the same value then Req = R/N. The total current gets divided in proportion to their individual conductance values. The current through the whole circuit is I = V Geq

IA

IR1

IR2

R1

R2

3A

G1= R11 G2 = R12

IR1

IR2

2Ω

1Ω

1 IR1 = I G1G+G 2

0.5 IR1 = 3 1+0.5 = 1A

2 IR2 = I G1G+G 2

1 = 2A IR2 = 3 1+0.5

(a) Fig. 1.6 Parallel circuit with a current source

(b)

5A

IR1

IR2

2Ω

3Ω

1/2 IR1 = 5 1/2+1/3 = 3A

1/3 = 2A IR2 = 5 1/2+1/3

(c)

8

1 Basic Concepts

and the current through any resistor Rn is IRn =

I Gn . Geq

In terms of resistance values, with just two resistors, Req = R1  R2 =

R1 R2 . R1 + R2

The parallel combination of two elements is denoted by the symbol . Consider the circuit shown in Fig. 1.6a. The circuit is energized by a current source of V = I A. The current through the two resistors are IR1 =

I G1 G1 + G2

and

I G2 . G1 + G2

IR2 =

Consider the circuit shown in Fig. 1.6b. The circuit is energized by a current source of I = 3 A. The current through the two resistors are IR1 =

3 0.5 = 1 A 1 + 0.5

and

3 1 = 2 A. 1 + 0.5

IR2 =

The voltage across both the resistors are the same, 1 × 2 = 2 × 1 = 2 V. The power consumed by the circuit is P = 2 × 3 = 6 W. Consider the circuit shown in Fig. 1.6c. The circuit is energized by a current source of I = 5 A. The current through the two resistors are IR1 =

5 = 3A 2(1/2 + 1/3)

and

IR2 =

5 = 2 A. 3(1/2 + 1/3)

As the source direction is reversed, the direction of current flow is also reversed. The voltage across both the resistors is the same, 3 × 2 = 2 × 3 = 6 V. The power consumed by the circuit is P = −5 × −6 = 30 W. Consider the circuit shown in Fig. 1.7a. The circuit is energized by a voltage source of V . The currents through the resistors are IR1 =

V R1

and

IR2 =

V . R2

Consider the circuit shown in Fig. 1.7b. The circuit is energized by a voltage source of 2 V. The currents through the resistors are IR1 =

2 = 1A 2

and

IR2 =

2 = 2 A. 1

1.4 Resistors Connected in Series and Parallel

V +

IR1

IR2

R1

R2

9

2V +

IR1

IR2

2Ω

1Ω

6V +

IR1

IR2

2Ω

3Ω

IR1 = RV1

IR1 = 22 = 1A

IR1 = 62 = 3A

IR2 = RV2

IR2 = 21 = 2A

IR2 = 63 = 2A

(a)

(b)

(c)

Fig. 1.7 Parallel circuit with a voltage source

The power consumed by the circuit is P = 2 × 3 = 6 W. Consider the circuit shown in Fig. 1.7c. The circuit is energized by a voltage source of 6 V. The currents through the resistors are IR1 =

6 = 3A 2

and

IR2 =

6 = 2 A. 3

As the polarity of the source is reversed, the direction of the current is also reversed. The power consumed by the circuit is P = −5 × −6 = 30 W.

1.4

Resistors Connected in Series and Parallel

The analysis of series and parallel circuits is relatively straightforward. In general, most circuits are a combination of series and parallel circuits or connected in a random configuration in which none of the elements is in series or parallel or a combination of both. Obviously, combinations of the concepts of series and parallel circuits are used to analyze series-parallel circuits. Analysis of circuits with random configurations is presented in later chapters. First, we have to identify the parts of the circuit with series and parallel configurations and simplify them separately. Now, the circuit gets reduced to a simpler form. These steps must be repeated until we can determine the source current. Then, using the voltage-division and current-division laws governing series and parallel circuits repeatedly, find the voltages and currents at all parts of the circuit. Consider the circuit shown in Fig. 1.8a. The circuit is energized by a current source of I = IR1 = 1 A. The source current flows through R1 and, therefore, IR1 = 1 A. The voltage across it is 1 × 2 = 2 V. The source current gets divided between R2 and R3 . Using current-division formula, we get IR2 =

3 I 1 1= A G2 = G2 + G3 1 + (1/3) 4

IR3 =

1 I 1 1 G3 = = A G2 + G3 1 + (1/3) 3 4

10

1 Basic Concepts

IR1 +

2Ω R1

1A

− IR2 +

IR1 +

+ IR3

R2 1Ω R3 −

3Ω

2Ω R1

1V

− IR2 +

R2 1Ω R3

−

−

IR1 = 1A, IR2 = 34 A, IR3 = 14 A

+ IR3 3Ω

−

11 1 4 Req = 2+ 1×3 4 = 4 Ω,IR1 = 11/4 = 11 A 3 1 3 1 IR2 = IR1 4 = 11 A, IR3 = IR1 4 = 11 A

(a)

(b)

Fig. 1.8 Resistors in series and parallel

as shown in the figure. 3 1 + = 1 A. 4 4 The voltage across the parallelly connected R2 and R3 is I = IR1 = IR2 + IR3 =

1 3 3 1 = 3 = V. 4 4 4 The power dissipated in the circuit is P = 12 × 2 +

3 × 1 = 2.75 W. 4

Consider the circuit shown in Fig. 1.8b. The circuit is energized by a voltage source of 1 V. First, we have to find the combined resistance of the circuit, which is Req = R1 + (R2  R3 ) =

11 . 4

Now, the current drawn from the source is I = IR1 =

4 1 = A. 11/4 11

This current gets divided between R2 and R3 . Using current-division formula, we get IR2 =

4 3 I 11 1= A G2 = G2 + G3 1 + (1/3) 11

IR3 =

4 1 1 I 11 = A G3 = G2 + G3 1 + (1/3) 3 11

1.5 Kirchhoff’s Voltage and Current Laws

11

as shown in the figure. 1 4 3 + = A 11 11 11 The voltage across the parallelly connected R2 and R3 is I = IR1 = IR2 + IR3 =

3 1 3 1= 3= V. 11 11 11 The combined voltage across the series-parallel combination is 2

4 3 11 + = = 1 V, 11 11 11

which is equal to the source voltage. The power dissipated in the circuit is P =

1.5

4 4 ×1= W. 11 11

Kirchhoff’s Voltage and Current Laws

Kirchhoff’s voltage and current laws are two of the few fundamental laws in electrical engineering. Voltage means electric potential difference. These laws are required to find the equilibrium of the circuit in terms of its currents and voltages. A circuit can be characterized by a set of independent variables, current or voltage. If currents are chosen as the variables, then Kirchhoff’s voltage law (KVL) is used to express the equilibrium of the circuit. The voltage at a point in the circuit is similar to the height of a point in a hilly terrain. Let us start at some point in a hill and climb the hill, visit the peaks, and climb down to the starting point. Then, the net height traversed by us is zero. Similarly, let us start at some point in a circuit, go through the circuit visiting junction points of elements and return back to the starting point. Then, the algebraic sum of the voltages around the loop is zero, which is KVL. Kirchhoff’s Voltage Law The algebraic sum of voltage drops across the circuit elements around a closed path of a circuit must be zero. That is,  ± V = 0. Consider the circuit shown in Fig. 1.9a. Let us start at point 1 and traverse the circuit in the clockwise direction. Voltage at point 1 minus the voltage at point 2 is the voltage drop from point 1 to 2, which is Fig. 1.9 Kirchhoff’s voltage and current laws

VR1

1 +

R1

V +

2 -

+

I1 N I3

R2 VR2

3−V + VR1 + VR2 = 0 (a)

I2 I1 + I2 − I3 = 0 (b)

12

1 Basic Concepts

positive. Similarly, the voltage drop from point 2 to 3 is also positive. However, the voltage drop from point 3 to 1 is negative. Therefore, some of the voltage drops must be negative and the rest positive so that the sum around the closed path (loop) is zero. In the circuit shown in Fig. 1.4b, we have 6 + 4 − 10 = 0 or 6 + 4 = 10 or − 6 − 4 + 10 = 0. In the circuit shown in Fig. 1.4c, we have −2 − 3 + 5 = 0 or 2 + 3 = 5 or 2 + 3 − 5 = 0. We can traverse the loop in the anticlockwise direction also. Kirchhoff’s Current Law (KCL) In a market, at the end of the day, the money spent by the customers must be equal to the money received by the merchants. In a traffic junction, the number of vehicles entering must be equal to the number of vehicles leaving. At the junction of water pipes, the inflow of water must be equal to the outflow. Similarly, at the junction of several branches of a circuit, the algebraic sum of the currents must be zero. That is the sum of the incoming currents is equal to the sum of the outgoing currents. The algebraic sum of currents flowing from branches towards a node in a circuit must be zero. That is,  ± I = 0. When voltages are chosen to characterize the circuit, the equilibrium of the circuit is expressed by the KCL. In Fig. 1.9b, applying KCL, we get I1 + I2 − I3 = 0 or I1 + I2 = I3 or − I1 − I2 + I3 = 0. The incoming currents must be assigned the opposite sign assigned to the outgoing currents. In Fig. 1.8a, 3 1 IR1 = IR2 + IR3 = + = 1 A. 4 4 In Fig. 1.8b, 1 4 3 IR1 = IR2 + IR3 = + = A. 11 11 11 Characterization of circuits on a current or voltage basis has a dual nature in that they are essentially similar with the roles of the current and voltage variables interchanged. The dual nature of the series and parallel circuits is shown in Table 1.1. Table 1.1 The dual nature of the series and parallel circuits

Parallel circuit

Series circuit

Geq = G1 + G2 I = I1 + I2 = V G1 + V G2 1 2 I1 = I G1G+G , I2 = I G1G+G 2 2

Req = R1 + R2 V = V1 + V 2 = I R 1 + I R 2 1 2 V1 = V R1R+R , V2 = V R1R+R 2 2

I1 = I, G1 → ∞,I2 = I, G2 → ∞ V1 = V , R1 → ∞,V2 = V , R2 → ∞

1.7 Summary

1.6

13

Applications

In almost all electric circuits used in applications, series and parallel connection of elements will occur in some parts of the circuit. Circuit elements, such as resistors, are readily available only at certain values. It is a common practice to connect them in series and/or parallel to find the element with the required value. Another common occurrence day-to-day is the use of electric cells in series in appliances, such as torchlight. Usually, all devices, such as a motor, fan, and bulb, are connected in series with a switch to put the device on or off. Our house wiring is in parallel, connecting group of devices to each line. This prevents the shutdown of the whole power supply, when a problem occurs in a certain device. In water heaters, the heater is connected with a thermostat, to stop and start the power supply as required to control the temperature of the water at the set level, and a switch. In electrical filters, which pass some frequency components of a signal while attenuating others, resistors, inductances, and capacitances are connected in series and/or parallel. In sensor circuits, a voltage is applied to the series connection of a standard resistor and a sensing resistor. The resistance of this resistor changes with some parameters such as pressure, temperature, etc. By measuring the change of the voltage drop across the standard resistor, the change in resistance of the sensor and, hence, the required parameter is measured. Airport runway lights are connected in series with a constant current through them. Therefore, the current, for a given voltage source, through them is kept low, requiring small conductors for the circuit. Certain mechanism is included so that the failure of one bulb will not prevent illuminating by the rest.

1.7

Summary

• Electrical and electronic engineering is indispensable in all applications of science and engineering, either in home, office, or industry. • Electricity is the movement of charged particles, such as electrons. • The flow of current is similar to the flow of water in a pipe. • A circuit is a closed loop, which allows the flow of charges from one place to another. Components in the circuit control the flow and use it to do work. • Charge is similar to the amount of water. Voltage is similar to water pressure. • Current is the number of electrons moving past a point in the circuit per second. Voltage is the pressure difference between two points, due to which the current flows. • The opposition to the flow of current is called the resistance. The resistance is 1  if 1 ampere current flows through it, when 1 V is applied across it. • The voltage across a resistor V is the current I flowing through it times its value R, known as Ohm’s law. • Resistors can be combined in a series and/or parallel or arbitrary configurations. • An electric circuit is an interconnection of components, such as battery, resistors, etc. • In a series circuit, same current flows through the circuit. • In a parallel circuit, same voltage is applied across all the elements in the circuit. • Kirchhoff’s voltage law states that the algebraic sum of voltage drops across the circuit elements around a closed path of a circuit must be zero. • Kirchhoff’s current law states that the algebraic sum of currents flowing from branches towards a node in a circuit must be zero.

14

1 Basic Concepts

• Characterization of circuits on a current or voltage basis has a dual nature in that they are essentially similar with the roles of the current and voltage variables interchanged. • Electric circuits are used in almost all applications of science and engineering.

Exercises 1.1 Find the current I and the voltages across the resistors, VR1 , VR2 , VR3 , VR4 , VR5 in the series circuit, shown in Fig. 1.10. Verify KVL. 1.1.1 V = 2 V, R1 = 3, R2 = 2, R3 = 4,

R4 = 1,

R5 = 5.

* 1.1.2 V = −3V,

R1 = 1,

R2 = 2,

R3 = 3,

R4 = 4,

R5 = 5.

V = −4 V,

R1 = 2,

R2 = 1,

R3 = 4,

R4 = 3,

R5 = 6.

1.1.3

1.2 Find the voltages across the resistors, VR1 , VR2 , VR3 , VR4 , VR5 in the series circuit, shown in Fig. 1.11. 1.2.1 I = 2 A, R1 = 3, R2 = 2,

R3 = 4,

R4 = 1,

R5 = 5.

1.2.2 I = −3 A,

R1 = 1,

R2 = 2,

R3 = 3,

R4 = 4,

R5 = 5.

I = −4A,

R1 = 2,

R2 = 1,

R3 = 4,

R4 = 3,

R5 = 6.

* 1.2.3 1.3 Find the voltage across the resistors, Fig. 1.10 Series circuit with a voltage source

+

VR1

+

VR2 R2

R1 V +

+ R3

VR5 R5

+

VR4 R4

+

I VR3

Exercises

15

Fig. 1.11 Series circuit with a current source

+

VR1

+

VR2 R2

R1 I

R3 VR5

+

VR4

R5 Fig. 1.12 Parallel circuit with a current source

VR3

+

R4

IR1

IR2

IR3

IR4

IR5

R1

R2

R3

R4

R5

I IR1

IR2

IR3

IR4

IR5

R2

R3

R4

R5

IA

Fig. 1.13 Parallel circuit with a voltage source

I

+

V +

R1

VR1 , VR2 , VR3 , VR4 , VR5 in the parallel circuit, shown in Fig. 1.12. Find the currents through the resistors, IR1 , IR2 , IR3 , IR4 , IR5 Verify that the sum of the currents through the resistors is equal to the source current. 1.3.1 I = 2 A, R1 = 3, R2 = 2, R3 = 4, R4 = 1, R5 = 5. 1.3.2 I = −3 A,

R1 = 1,

R2 = 2,

R3 = 3,

R4 = 4,

R5 = 5.

I = −4 A,

R1 = 2,

R2 = 1,

R3 = 4,

R4 = 3,

R5 = 6.

* 1.3.3

1.4 Find the current I and the voltages across the resistors, VR1 , VR2 , VR3 , VR4 , VR5 in the parallel circuit, shown in Fig. 1.13. Find the currents through the resistors, IR1 , IR2 , IR3 , IR4 , IR5 Verify that the sum of the currents through the resistors is equal to the total current, I .

16

1 Basic Concepts

1.4.1 V = 2 V,

R1 = 3,

R2 = 2,

R3 = 4,

R4 = 1,

R5 = 5

* 1.4.2 V = −3 V,

R1 = 1,

R2 = 2,

R3 = 3,

R4 = 4,

R5 = 5

V = −4 V,

R1 = 2,

R2 = 1,

R3 = 4,

R4 = 3,

R5 = 6

1.4.3

1.5 Find the current I and the voltages across the resistors, VR1 , VR2 , VR3 , VR4 , VR5 and current through IR1 , IR2 , IR3 , IR4 , IR5 in the series-parallel circuit, shown in Fig. 1.14. Verify KVL around the loops and KCL at node x. 1.5.1 V = 2 V, R1 = 3, R2 = 2, R3 = 4, R4 = 1, R5 = 5. * 1.5.2 V = −3 V,

R1 = 1,

R2 = 2,

R3 = 3,

R4 = 4,

R5 = 5.

V = −4 V,

R1 = 2,

R2 = 1,

R3 = 4,

R4 = 3,

R5 = 6.

1.5.3

1.6 Find the voltages across the resistors, VR1 , VR2 , VR3 , VR4 , VR5 in the series-parallel circuit, shown in Fig. 1.15. Verify KCL at node x. 1.6.1 I = 2 A, R1 = 3, R2 = 2, R3 = 4, R4 = 1,

Fig. 1.14 Series-parallel circuit with a voltage source

R5 = 5.

R5 V +

IR1

IR2

xI IR3 IR4

R1

R2

R3

R4

Exercises

17

Fig. 1.15 Series-parallel circuit with a current source

I

IR1

R5 xI IR2 IR3 IR4

R1

R2

R3

R4

* 1.6.2 I = −3 A,

R1 = 1,

R2 = 2,

R3 = 3,

R4 = 4,

R5 = 5.

I = −4 A,

R1 = 2,

R2 = 1,

R3 = 4,

R4 = 3,

R5 = 6.

1.6.3

DC Circuits

2

An electric circuit, for theoretical analysis, is an interconnection of idealized representation of physical components, such as voltage and current sources, switches, resistors, inductors, and capacitors. Any physical device can be approximated for practical purposes by idealized devices with sufficient accuracy. The relationship between the voltage across an element and the current through it is called its volt–ampere relationship and is assumed to be linear in the specified operating ranges. In this chapter, we consider circuits with DC sources and resistors only. DC is abbreviated form for direct current, a common designation for constant voltage or current. AC is abbreviated form for alternating current, a common designation for sinusoidally varying voltage or current. Apart from its practical importance, the study of the basic principles of DC circuit analysis, without considering the transient response, is relatively simpler. Further, it has the priority since the mathematical process of computing the circuit response for a constant excitation is relatively easier to learn. The same basic principles can be easily adapted for AC circuit analysis. Therefore, a good understanding of the DC circuit analysis is essential and it makes the learning of the AC circuit analysis that much simpler. With the sources and circuit elements specified, the purpose of circuit analysis is to determine the voltages and currents at all parts of the circuit. There is a voltage across any circuit element, called a branch of the circuit, and a current through it. Since the current and voltage in an element are related through its volt–ampere relationship, circuits can be analyzed in terms of the branch currents alone or branch voltages alone. A minimum set of independent branch voltages or currents has to be determined. There is no unique choice. The method of circuit analysis based on currents is called mesh or loop analysis. The other method based on voltages is called nodal analysis. Either method results in a set of equilibrium equations. The solution to these equations is obtained using linear algebra, yielding the values of the independent variables. From these values, all the values of other branch voltages and currents of the circuit can be determined, using KVL and KCL, completing the circuit analysis. The analysis presented, hereafter, is an extension of the analysis presented in the first chapter applicable to circuits with random configurations. We use the same input–output relationship of elements, Ohm’s law, KVL, and KCL, but in a systematic manner as the circuit complexity is high. The equilibrium conditions for a circuit can be established in either of the two ways:

© Springer Nature Switzerland AG 2020 D. Sundararajan, Introductory Circuit Theory, https://doi.org/10.1007/978-3-030-31985-4_2

19

20

2 DC Circuits

Fig. 2.1 (a) Ideal and (b) practical voltage sources and their volt–ampere characteristics

Rs I

I + V −

Vt Vt V

+ V −

V

(a)

I

(b)

I

1. Through a set of N equations, using KVL, in which the mesh currents are the independent variables. 2. Through a set of M equations, using KCL, in which the branch voltages are the independent variables.

Voltage Sources A source is required to energize a circuit. An ideal voltage source is characterized by its volt–ampere relationship of keeping its terminal voltage same irrespective of the current drawn by the load circuit connected to it, as shown in Fig. 2.1a. A practical voltage source can be approximated by an ideal voltage source in series with a resistance Rs , called the source resistance. Therefore, the voltage at the terminals of a practical voltage source, Vt , will decrease as the current drawn from it is increased, as shown in Fig. 2.1b. A voltage source is an applied constraint in the circuit. Due to this constraint, the voltages at various nodes of the circuit are fixed, depending on the characteristics of the voltage source and the circuit. The + sign or an arrow indicates that a voltage raise in that direction is considered as a positive quantity. Two voltage sources with capacities V1 and V2 volts, connected in series, are equivalent to a single source with capacity, which is equal to the algebraic sum of V1 and V2 . Two voltage sources with unequal voltage capacities are not allowed to be connected in parallel, since it is a violation of KVL.

2.1

Mesh Analysis

Mesh analysis is also called loop analysis. The response of a circuit is completely characterized by the values of currents and voltages in all its branches. A circuit, geometrically, is characterized by its branches and nodes. A branch represents a single element, such as a resistor. That is, a branch is any two-terminal element. However, in certain problems, it is possible to regard series, parallel, or seriesparallel combination of elements as a branch. The point where the terminals of two or more branches are connected together is called a node. Nodes, with selected independent voltages, are indicated by a dot in the circuit diagram. If two nodes are connected by a conductor (a short circuit), then the two nodes constitute a single node. A loop is a closed path in a circuit. It starts at a node, passes through a set of nodes (passing through each node only once), and returns to the starting node. A loop is independent if it contains at least one branch that is not part of any other independent loop. Two or more elements are connected in series if they carry the same current. Two or more elements are connected in parallel if their terminal voltages are the same. A tree is any set of branches of the circuit that is sufficient to connect all its nodes. As the structure of the selected set of branches resembles that of a tree, it is called a tree. The number of branches forming a tree is TB = N − 1, where TB is the number of branches of the tree, called tree branches, and N is the number of nodes of the circuit. A tree contains no closed paths. There are several possible

2.1 Mesh Analysis

21

I1 = 0.6A I2 = 0.2A

Fig. 2.2 (a) A bridge circuit; (b) the circuit with branch currents

R1

1Ω

R2 R5

1V +

R1

3Ω 1V

1Ω R3

3Ω

(a)

R4

1Ω

R3

0.4A 1Ω R2 R5 1 Ω, 0.2A 3Ω R4 0.2A I3 = 0.4A

3Ω

1Ω

(b)

different trees for a given circuit. There is no unique tree corresponding to a circuit. But, the number of tree branches is always TB. Any one of the possible trees is sufficient for analyzing the corresponding circuit. Branches removed from a circuit, in forming a tree, are called the links. The number of links is M. Then, the total number of branches, B, in the circuit is B = TB + M. The equilibrium state of a circuit is determined by L independent link currents flowing through the L selected loops. That is, L branch currents only, of the B branch currents, are independent and it is the smallest number of currents in terms of which the rest can be expressed uniquely as a linear combination, using KCL at the nodes. Given a circuit, the problem is to determine all the voltages across the branches and currents through the branches. The branch voltages are related to branch currents through volt–ampere characteristics of the particular circuit elements. Therefore, the behavior of any circuit is adequately characterized by branch currents alone or branch voltages alone. Based on this, there are two basic methods of circuit analysis: (1) mesh or loop analysis and (2) nodal analysis. In mesh analysis, the values of the independent current variables are determined. In nodal analysis, the values of the independent voltage variables are determined. In order to reduce mistakes and confusion, the problem has to be approached systematically through several steps. Steps in establishing equilibrium equations for the mesh analysis of a circuit and finding the solution are as follows: 1. Select an appropriate number of independent current variables and the directions of current flow 2. Express the dependent current variables, by applying KCL at nodes, in terms of independent current variables 3. Apply KVL around the selected loops to set up a set of simultaneous equations 4. Solve for the independent currents and find the currents in all the branches 5. Verify the solution using KVL and KCL The concepts are better presented through examples. Consider the analysis of the bridge circuit, shown in Fig. 2.2a. The circuit consists of 5 resistors with their values, designated R1 to R5 , and a 1 V voltage source. Each circuit is different in the kinds of elements involved and the way they are interconnected. A resistor is characterized by its volt–ampere relationship of the voltage across it being proportional to the current through it, the constant of proportionality being its resistance value in ohms. One possible set of independent current variables is shown in Fig. 2.2b. There are three links in the circuit and, therefore, the number of independent current variables is three, designated as

22

2 DC Circuits

I1

R1

R1

1Ω R5

I2

1Ω

R1

1Ω

3Ω

3Ω R5

R5

1V

1Ω R3

R2

1Ω

1Ω R3

3Ω

R3

3Ω

R4

1Ω

I3 Fig. 2.3 A tree, with four nodes shown by discs, corresponding to the bridge circuit and the three independent loops

{I1 , I2 , I3 }. The tree corresponding to the circuit, with four nodes shown by discs, is shown on the left side in Fig. 2.3, followed by the three loops formed by inserting the three links, in turn, in the tree. The three links are the voltage source and resistors R2 and R4 . The three branches pertaining to the three independent currents must be the links associated with the selected tree. The insertion of one of the links in the tree must result in only one closed loop, which is different for each link. Therefore, a distinct set of loops is obtained. Step 1 Independent and Dependent Variables The three independent current variables are I1 , I2 , and I3 . By applying KCL at three nodes in Fig. 2.2b, the three dependent currents, flowing through resistors R1 , R3 , and R5 , are expressed in terms of the independent currents as IR1 = I1 − I2 IR3 = I1 − I3 IR5 = IR1 − IR3 = I3 − I2 . The directions of all the branch currents, which can be arbitrary, must be assigned. If the assumed direction of the current is not correct, then the value of the current found in the analysis will be negative. Step 2 Setting Up the Equilibrium Equations The equilibrium conditions for a circuit can be expressed, in mesh analysis, by a set of N equations using KVL in which the N loop currents are the variables. Assuming no current sources and N independent loops, the equilibrium equations, in general, are of the form Z11 I1 + Z12 I2 + Z13 I3 + · · · + Z1N IN = Z21 I1 + Z22 I2 + Z23 I3 + · · · + Z2N IN = ···

 

V1 V2

= ···  ZN 1 I1 + ZN 2 I2 + ZN 3 I3 + · · · + ZN N IN = VN ,

2.1 Mesh Analysis

23

where {I1 , I2 , . . . , IN } are the currents in the N chosen loops of the circuit, {Zi1 , Zi2 , . . . , ZiN } are    the total impedances in the respective loops and { V1 , V2 , . . . , VN } are the algebraic sum of the source voltages in the respective loops. For example, Z12 is the impedance in the branch common to loops 1 and 2. It is positive or negative depending on the directions of the flow of the currents in the common branch. Impedance has a resistive part and a reactive part. In this chapter, we use only resistors. Therefore, Z = R. The equilibrium equations hold whether the circuit consists of resistors only or otherwise. In order to emphasize this fact and to make the understanding of the analysis of AC circuits easier, we use Z mostly instead of R. Using vector and matrix quantities, the equilibrium equations and the solution are concisely given by ZI = V and I = Z −1 V , where Z is the impedance matrix, I is the current vector, and V is the voltage vector. ⎡

Z11 Z12 ⎢ Z21 Z22 Z=⎢ ⎣ ZN 1 ZN 2

Z13 Z23 ··· ZN 3

⎤

· · · Z1N · · · Z2N ⎥ ⎥, ⎦ ··· · · · ZN N

⎡

I1 ⎢ I2 ⎢ ⎢ I = ⎢ I3 ⎢ . ⎣ .. IN

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎦

⎡ ⎤ V  1 ⎢ V2 ⎥ ⎢ ⎥ ⎢ ⎥ V = ⎢ V3 ⎥ . ⎢ . ⎥ ⎣ .. ⎦  VN

For the circuit shown in Fig. 2.2b, applying KVL around the chosen loops, we get the three equilibrium equations. Z1 (I1 − I2 ) + Z3 (I1 − I3 ) = 1 −Z1 (I1 − I2 ) − Z5 (I3 − I2 ) + Z2 I2 = 0 −Z3 (I1 − I3 ) + Z5 (I3 − I2 ) + Z4 I3 = 0. These are the equations obtained traversing each loop either in the clockwise or anticlockwise direction. For example, the first equation corresponds to the loop with current I1 . Considering the direction of the current flow, the voltage drops across R1 and R3 add up, whereas the source voltage polarity is opposite of this drop. Therefore, IR1 Z1 + IR3 Z3 = Z1 (I1 − I2 ) + Z3 (I1 − I3 ) = 1. This is a straightforward application of KVL to loop 1. There are no sources in the other loops. Simplifying the equilibrium equations, we get (Z1 + Z3 )I1 − Z1 I2 − Z3 I3 = 1 −Z1 I1 + (Z1 + Z2 + Z5 )I2 − Z5 I3 = 0 −Z3 I1 − Z5 I2 + (Z3 + Z4 + Z5 )I3 = 0. Using matrices, we get ⎤⎡ ⎤ ⎡ ⎤ 1 I1 −Z1 −Z3 (Z1 + Z3 ) ⎣ −Z1 (Z1 + Z2 + Z5 ) −Z5 ⎦ ⎣ I2 ⎦ = ⎣ 0 ⎦ . 0 −Z3 −Z5 (Z3 + Z4 + Z5 ) I3 ⎡

24

2 DC Circuits

Ensure that the correct values for the impedances and sources and the appropriate equations are used before proceeding with any numerical computation. With Z1 = 1, we get

Z2 = 3,

Z3 = 3,

Z4 = 1 and Z5 = 1,

⎤⎡ ⎤ ⎡ ⎤ 1 4 −1 −3 I1 ⎣ −1 5 −1 ⎦ ⎣ I2 ⎦ = ⎣ 0 ⎦ . I3 0 −3 −1 5 ⎡

The determinant of the impedance matrix must be nonzero. Otherwise, the equilibrium equations are dependent and must be checked. The determinant is 40 for this example. Step 3 Solving the Equilibrium Equations In order to solve for currents, we have to find the inverse of the impedance matrix. For this example, the inverse is found, as shown in Appendix A. In order to verify the inverse, it must be multiplied by the impedance matrix to get a 3 × 3 identity matrix, a matrix with all zeros except 1 on the main diagonal. I = Z −1 V ⎤ ⎤⎡ ⎤ ⎡ ⎤−1 ⎡ ⎤ ⎡ ⎡ ⎤ ⎡ 0.6 1 0.6000 0.2000 0.4000 4 −1 −3 1 I1 ⎣ I2 ⎦ = ⎣ −1 5 −1 ⎦ ⎣ 0 ⎦ = ⎣ 0.2000 0.2750 0.1750 ⎦ ⎣ 0 ⎦ = ⎣ 0.2 ⎦ I3 0.4 0 0.4000 0.1750 0.4750 −3 −1 5 0 {IR1 = I1 − I2 = 0.4 A,

IR3 = I1 − I3 = 0.2 A,

IR5 = I3 − I2 = 0.2 A}.

All the currents are shown in Fig. 2.2b. Verifying the Solutions Applying KVL to each loop, in turn, we get IR1 R1 + IR3 R3 = (0.4)1 + (0.2)3 = 1 −I2 R2 + IR5 R5 + IR1 R1 = −(0.2)3 + (0.2)1 + (0.4)1 = 0 IR5 R5 + I3 R4 − IR3 R3 = (0.2)1 + (0.4)1 − (0.2)3 = 0. Verifying the solution using KCL at the four nodes I1 − IR1 − I2 = 0.6 − 0.4 − 0.2 = 0 IR1 − IR3 − IR5 = 0.4 − 0.2 − 0.2 = 0 IR5 + I2 − I3 = 0.2 + 0.2 − 0.4 = 0 IR3 − I1 + I3 = 0.2 − 0.6 + 0.4 = 0. Symmetry of the Impedance Matrix The procedure for loop analysis described is applicable for any circuit. However, for certain type of circuits under some conditions, the general procedure can be simplified. Of course, if a shortcut is available for the analysis of circuits it should be taken advantage of. The impedance matrix of the bridge circuit is symmetrical about its principal diagonal. That is, Zij = Zj i . It is useful to verify the setting of the equilibrium equations. It is not an inherent property of linear circuits. This

2.1 Mesh Analysis

25

is due to the setting of KVL equations using the same loops defining the link currents and using consistently clockwise or anticlockwise direction for the independent currents in a mappable circuit. The graph of a mappable circuit does not have any branches crossing each other. Further, all the voltage sources in the circuit must be independent. In the first loop of the example circuit, shown in Fig. 2.3, the independent current I1 flows in the same direction through all the elements. Therefore, the contribution of the voltage due to I1 is the current multiplied by the sum of the values of all resistors in the first loop. Consequently, the first value in the main diagonal of the impedance matrix is 4. Similarly, the other values in the main diagonal are, respectively, the sum of the values of the resistors in the corresponding loops. The second and third values are 5 and 5. The other entries in the matrix are negative and symmetric. Their values are the values of the resistance common to the two loops. For example, the value of the resistor between the first and third loops is 3 and, therefore, z13 = z31 = −3. That is, the voltage drop in loop 1 by the current I3 is −3 V, since it flows in the opposite direction of that of I1 . In the other cases, the values are −1. If an independent current does not flow in a certain loop, then the entry in the matrix is zero. It is possible to use two different sets of independent loops for the independent currents and the KVL equations. Then, the symmetry may not exist. Further, the determinant of the impedance matrix should be nonzero in any case.

Changed Assumption of the Current Directions It is better to choose the current directions to be consistently clockwise or counterclockwise. However, the selection of the direction of the currents in the branches can be arbitrary. In that case, of course, the symmetry of the impedance matrix may be lost. If a current flows in the direction opposite to that assumed, the analysis result will be negative-valued. Let us do the analysis of the bridge circuit again with different direction for the currents, as shown in Fig. 2.4. Applying KCL at the nodes, we get IR1 = −(I1 + I2 ) IR3 = I3 − I1 IR5 = I3 + I2 . The equilibrium equations are −Z1 (I1 + I2 ) − Z3 (I1 − I3 ) = 1 Z1 (I1 + I2 ) + Z5 (I2 + I3 ) + Z2 I2 = 0 Z3 (I3 − I1 ) + Z5 (I2 + I3 ) + Z4 I3 = 0. I1 = −0.6A I2 = 0.2A

Fig. 2.4 The bridge circuit with different current directions

R1

1Ω

R2

3Ω

R5

1V

R1

0.4A 1Ω R2 R5

R3

1 Ω, −0.2A 3Ω R4 1 Ω 0.2A I3 = −0.4A

1V

1Ω R3

3Ω

R4

1Ω

3Ω

26

2 DC Circuits

Simplifying, we get −(Z1 + Z3 )I1 − Z1 I2 + Z3 I3 = 1 Z1 I1 + (Z1 + Z2 + Z5 )I2 + Z5 I3 = 0 −Z3 I1 + Z5 I2 + (Z3 + Z4 + Z5 )I3 = 0. In matrix form, ⎤⎡ ⎤ ⎡ ⎤ 1 I1 −Z1 Z3 −(Z1 + Z3 ) ⎣ Z1 (Z1 + Z2 + Z5 ) Z5 ⎦ ⎣ I2 ⎦ = ⎣ 0 ⎦ . 0 −Z3 Z5 (Z3 + Z4 + Z5 ) I3 ⎡

With Z1 = 1,

Z2 = 3,

Z3 = 3,

Z4 = 1,

Z5 = 1

⎤⎡ ⎤ ⎡ ⎤ I1 −4 −1 3 1 ⎣ 1 5 1 ⎦ ⎣ I2 ⎦ = ⎣ 0 ⎦ . −3 1 5 I3 0 ⎡

Solving the equilibrium equations, we get ⎤ ⎤⎡ ⎤ ⎡ ⎤−1 ⎡ ⎤ ⎡ ⎤ ⎡ −0.6 1 −0.6000 −0.2000 0.4000 −4 −1 3 1 I1 ⎣ I2 ⎦ = ⎣ 1 5 1 ⎦ ⎣ 0 ⎦ = ⎣ 0.2000 0.2750 −0.1750 ⎦ ⎣ 0 ⎦ = ⎣ 0.2 ⎦ I3 −0.4 0 −0.4000 −0.1750 0.4750 −3 1 5 0 ⎡

{IR1 = 0.4,

IR3 = 0.2,

IR5 = −0.2}.

A negative value for the current indicates current flow in the opposite direction from that assumed. After correcting the directions, we get the direction shown in Fig. 2.2b. The power consumed by the circuit can be computed by multiplying both sides of the equilibrium equations, by the currents {I1 , I1 , I3 , . . . , IN , } respectively, and summing I1 (Z11 I1 + Z12 I2 + Z13 I3 + · · · + Z1N IN ) = I2 (Z21 I1 + Z22 I2 + Z23 I3 + · · · + Z2N IN ) =

 

V1 I1 V2 I2

···

= ···  IN (ZN 1 I1 + ZN 2 I2 + ZN 3 I3 + · · · + ZN N IN ) = VN IN . For the specific example, we get ⎤ ⎤ ⎡ ⎤⎡ 1(0.6) 0.6 0.6(4 −1 −3) ⎣ 0.2(−1 5 −1) ⎦ ⎣ 0.2 ⎦ = ⎣ 0⎦. 0 0.4 0.4(−3 −1 5) ⎡

Summing either side yields the power consumed by the circuit as 0.6 W. This procedure is useful to compute the power with multiple sources.

2.2 Nodal Analysis

2.2

27

Nodal Analysis

The equilibrium state of a circuit can also be determined by TB independent voltages, where TB is the number of branches of the circuit forming a tree. KCL is applied at each node (except one, which becomes the ground node) of the selected tree of the circuit to set up the required equilibrium equations. The steps in nodal analysis are as follows: 1. Select an appropriate number of independent voltage variables 2. Express the dependent voltage variables, by applying KVL around the loops, in terms of independent voltage variables 3. Apply KCL at the selected nodes to set up a set of simultaneous equations 4. Solve for the independent voltages and find the voltages at all the nodes 5. Verify the solution using KVL and KCL Prefer the node connected to the maximum number of elements and sources as the ground node. A ground node acts as a reference for voltage levels at various points in the circuit. The voltage at the ground node is assumed to be zero. Let us analyze the same bridge circuit, shown again in Fig. 2.5a, by nodal analysis. Step 1 Independent and Dependent Variables The nodes are shown in Fig. 2.5b. The three independent voltages are V1 , V2 , and V3 . By applying KVL around the three loops in Fig. 2.5b, the other three dependent voltages, across the resistors R1 , R2 , and R5 , are expressed in terms of the independent voltages as VR1 = V1 − V2 VR2 = V1 − V3 VR5 = V2 − V3 . For example, applying KVL to the top right loop, we get VR1 + (V2 − V3 ) + (V3 − V1 ) = 0 and

VR1 = V1 − V2 .

Fig. 2.5 Nodal analysis of the bridge circuit

V1 = 1V

R1

1Ω

R2 R5

1V

1V

1Ω R3

3Ω

(a)

R1

3Ω

R4

1Ω

0.4V 1Ω R2 R5

3Ω

V2 V3 = 0.4V 0.6V 1 Ω, 0.2 V R4 1 Ω R3 3 Ω

(b)

28

2 DC Circuits

Step 2 Setting Up the Equilibrium Equations Assuming no voltage sources and N independent nodes, applying KCL at each node in turn, the equilibrium equations, in general, are of the form Y11 V1 + Y12 V2 + Y13 V3 + · · · + Y1N VN = Y21 V1 + Y22 V2 + Y23 V3 + · · · + Y2N VN =

 

I1 I2

··· = ···  YN 1 V1 + YN 2 V2 + YN 3 V3 + · · · + YN N VN = IN , where {V1 , V2 , . . . , VN } are the voltages at the chosen N nodes of the circuit with respect to the ground node, {Yi1 , Yi2 , . . . , YiN } are the admittances in the branches joining nodes i and j , and    { I1 , I2 , . . . , IN } are the algebraic sum of the currents from current sources, connected to the ground node, feeding the respective nodes. Admittance has a conductive part and a susceptive part. In this chapter, we use only resistors. Therefore, Y = G. The equilibrium equations hold whether the circuit consists of resistors only or otherwise. In order to emphasize this fact and to make the understanding of the analysis of AC circuits easier, we use Y instead of G. Using vector and matrix quantities, the equilibrium equations and the solution are concisely given by Y V = I and V = Y −1 I , where V is the voltage vector, Y is the admittance matrix, and I is the current vector. ⎡

Y11 Y12 ⎢ Y21 Y22 Y =⎢ ⎣ YN 1 YN 2

Y13 Y23 ··· YN 3

⎤

· · · Y1N · · · Y2N ⎥ ⎥, ⎦ ··· · · · YN N

⎡

V1 ⎢ V2 ⎢ ⎢ V = ⎢ V3 ⎢ . ⎣ ..

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎦

VN

⎡ ⎤ I  1 ⎢ I2 ⎥ ⎢ ⎥ ⎢ ⎥ I = ⎢ I3 ⎥ . ⎢ . ⎥ ⎣ .. ⎦  IN

Let us solve for the node voltages for the circuit shown in Fig. 2.5. Since V1 = 1 is known, there are only 2 unknowns, {V2 , V3 }. When a voltage source is connected between a node and the ground node, the problem is simplified. With Z = R and applying KCL at nodes with voltages V2 and V3 , we get the equilibrium equations, respectively, at nodes 2 and 3 as (V2 − 1) (V2 − V3 ) V2 + + =0 Z1 Z5 Z3 (V3 − 1) (V3 − V2 ) V3 + + = 0. Z2 Z5 Z4 In loop analysis, we can traverse the loop in the clockwise or counterclockwise direction. Similarly, in nodal analysis, we can write the KCL equations at each node so that the currents are directed out of the node or flowing towards the node. For this example, the KCL equations at each node are written so that the currents are directed out of the node. Simplifying the equilibrium equations, we get (Z3 Z1 + Z1 Z5 + Z3 Z5 )V2 − Z3 Z1 V3 = Z3 Z5 −Z2 Z4 V2 + (Z2 Z4 + Z2 Z5 + Z4 Z5 )V3 = Z4 Z5 .

2.3 Examples

29

With Z1 = 1,

Z2 = 3,

Z3 = 3,

Z4 = 1,

Z5 = 1

7V2 − 3V3 = 3 −3V2 + 7V3 = 1.

In matrix form

7 −3 −3 7



V2 V3



 3 = . 1

The determinant of the admittance matrix is 40. Step 3 Solving the Equilibrium Equations V = Y −1 I The inverse of an arbitrary 2 × 2 matrix A exists if |A| = (a11 a22 − a12 a21 ) = 0. Then, A−1 is given by

a a A = 11 12 a21 a22

V2 V3







−1

and

7 −3 = −3 7

A



1 a22 −a12 = (a11 a22 − a12 a21 ) −a21 a11

−1 

 

 3 0.1750 0.0750 3 0.6 = = 1 0.0750 0.1750 1 0.4

{VR1 = 0.4,

VR2 = 0.6,

VR5 = 0.2}.

These values are the same as those obtained by loop analysis.

2.3

Examples

A bridge circuit, with the voltage source in the middle, is shown in Fig. 2.6b. The tree corresponding to the circuit has 4 nodes. With the polarities marked same as in Fig. 2.2b, three independent currents

V1 = 0.125V 0.25A I1 0.125A R1 R6

1Ω

I2 = 0.125A R2

I1

I2

3Ω

R5

0.75V 1Ω V2 V3 R 6 1V 0.375V 1Ω 0.375A −0.25V R4 1Ω R3 3Ω 0.125A I3 = 0.25A (a)

Fig. 2.6 A bridge circuit, with the voltage source in the middle

R1

1Ω

R2

3Ω

R5

1Ω

1V

1Ω R3

3Ω

R4 I3 (b)

1Ω

30

2 DC Circuits

V1 I1 R1

1Ω R5

V2 R3

V3

1V

1Ω

R6

3Ω

(a)

R1

1Ω

R3

3Ω

1Ω

(b) R5

I2 R1

1Ω

R2

3Ω

R5

1V

1Ω R3

3Ω

R4 I3

1V

1Ω

1Ω

(c)

(d)

Fig. 2.7 A tree, with four nodes shown by discs, corresponding to the bridge circuit and the three independent loops

are to be found using the mesh analysis. The equilibrium equations are similar to that for the circuit in Fig. 2.2. The tree corresponding to the circuit, with four nodes shown by discs, is shown in Fig. 2.7a. Figure 2.7b–d show the three loops formed by inserting the three links, in turn, in the tree. The three links are the resistors R6 , R2 , and R4 . The three independent current variables are I1 , I2 and I3 . By applying KCL at three nodes in Fig. 2.6b, the other three dependent currents, flowing through resistors R1 , R3 and R5 , are expressed in terms of the independent currents as IR1 = I1 − I2 IR3 = I1 − I3 IR5 = I3 − I2 . The equilibrium equations are, respectively, for the loops corresponding to the independent currents, I1 , I2 , and I3 . Z1 (I1 − I2 ) + Z3 (I1 − I3 ) + Z6 I1 = 0 −Z1 (I1 − I2 ) − Z5 (I3 − I2 ) + Z2 I2 = 1 −Z3 (I1 − I3 ) + Z5 (I3 − I2 ) + Z4 I3 = −1. Simplifying, the equations in matrix form are ⎡ ⎤⎡ ⎤ ⎡ ⎤ (Z1 + Z3 + Z6 ) I1 0 −Z1 −Z3 ⎣ −Z1 (Z1 + Z2 + Z5 ) −Z5 ⎦ ⎣ I2 ⎦ = ⎣ 1 ⎦ . −Z3 −Z5 (Z3 + Z4 + Z5 ) I3 −1 With Z1 = 1,

Z2 = 3,

Z3 = 3,

Z4 = 1,

Z5 = 1,

Z6 = 1,

2.3 Examples

31

we get 5I1 − I2 − 3I3 = 0 −I1 + 5I2 − I3 = 1 −3I1 − I2 + 5I3 = −1. In matrix form,

⎤ ⎤⎡ ⎤ ⎡ 0 5 −1 −3 I1 ⎣ −1 5 −1 ⎦ ⎣ I2 ⎦ = ⎣ 1 ⎦ . I3 −1 −3 −1 5 ⎡

For this circuit, we can get the impedance matrix from inspection. The determinant of the impedance matrix is 64. Solving for the currents, we get ⎤ ⎤ ⎡ ⎤⎡ ⎤−1 ⎡ ⎤ ⎡ ⎤ ⎡ −0.125 0 0.375 0.125 0.250 5 −1 −3 0 I1 ⎣ I2 ⎦ = ⎣ −1 5 −1 ⎦ ⎣ 1 ⎦ = ⎣ 0.125 0.250 0.125 ⎦ ⎣ 1 ⎦ = ⎣ 0.125 ⎦ I3 −0.250 −1 0.250 0.125 0.375 −3 −1 5 −1 ⎡

{IR1 = −0.250,

IR3 = 0.125,

IR5 = −0.375}.

Some of the currents happen to flow in the reverse direction, indicated by a negative value, to that with the initial assumption. Figure 2.6a shows the circuit with correct current directions and the node voltages. From inspection, current and voltage values satisfy KCL and KVL. For example, at the top left node, the incoming current is 0.25 and the sum of the outgoing currents is 0.125 + 0.125 = 0.25. For the left most loop, the sum of the voltage drops across resistors R1 and R6 is 0.375 and the drop across R3 is also 0.375 with the polarities reversed. Let us do the nodal analysis for the circuit. The equilibrium equations are (V1 − V2 ) V1 (V1 − V3 ) + + =0 Z1 Z6 Z2 (V2 − V1 ) V2 (V2 − (V3 + 1)) + + =0 Z1 Z3 Z5 V3 (V3 − V1 ) ((V3 + 1) − V2 ) + + = 0. Z4 Z2 Z5 Simplifying, we get (Z2 Z6 + Z1 Z2 + Z1 Z6 )V1 − Z2 Z6 V2 − Z1 Z6 V3 = 0 −Z3 Z5 V1 + (Z3 Z5 + Z1 Z5 + Z1 Z3 )V2 − Z1 Z3 V3 = Z3 Z1 −Z4 Z5 V1 − Z2 Z4 V2 + (Z2 Z5 + Z4 Z5 + Z4 Z2 )V3 = −Z2 Z4 . With Z1 = 1,

Z2 = 3,

Z3 = 3,

Z4 = 1,

Z5 = 1, Z6 = 1,

the equilibrium equations, in matrix equation form, are ⎤⎡ ⎤ ⎡ ⎤ V1 7 −3 −1 0 ⎣ −3 7 −3 ⎦ ⎣ V2 ⎦ = ⎣ 3 ⎦ . −1 −3 7 V3 −3 ⎡

32

2 DC Circuits

The determinant of the admittance matrix is 192. Solving for the voltages, we get ⎤ ⎤ ⎡ ⎤⎡ ⎤−1 ⎡ ⎤ ⎡ ⎤ ⎡ 0.125 0 0.2083 0.1250 0.0833 7 −3 −1 0 V1 ⎣ V2 ⎦ = ⎣ −3 7 −3 ⎦ ⎣ 3 ⎦ = ⎣ 0.1250 0.2500 0.1250 ⎦ ⎣ 3 ⎦ = ⎣ 0.375 ⎦ V3 −0.250 −3 0.0833 0.1250 0.2083 −1 −3 7 −3 ⎡

{VR1 = −0.250,

VR2 = 0.375,

VR5 = −0.375}.

Let us analyze the circuit with correct current directions. IR1 = I1 + I2 IR3 = I3 − I1 IR5 = I3 + I2 . The equilibrium equations are Z1 (I1 + I2 ) + Z3 (I1 − I3 ) + Z6 I1 = 0 Z1 (I1 + I2 ) + Z5 (I3 + I2 ) + Z2 I2 = 1 Z3 (I3 − I1 ) + Z5 (I3 + I2 ) + Z4 I3 = 1. Simplifying, we get (Z1 + Z3 + Z6 )I1 + Z1 I2 − Z3 I3 = 0 Z1 I1 + (Z1 + Z2 + Z5 )I2 + Z5 I3 = 1 −Z3 I1 + Z5 I2 + (Z3 + Z4 + Z5 )I3 = 1. With Z1 = 1,

Z2 = 3,

Z3 = 3,

Z4 = 1,

Z5 = 1,

Z6 = 1,

V = 1,

we get 5I1 + I2 − 3I3 = 0 I1 + 5I2 + I3 = 1 −3I1 + I2 + 5I3 = 1. In matrix form,

⎤⎡ ⎤ ⎡ ⎤ 0 I1 5 1 −3 ⎣ 1 5 1 ⎦ ⎣ I2 ⎦ = ⎣ 1 ⎦ . I3 1 −3 1 5 ⎡

Solving for the currents, we get ⎡

⎤ ⎤⎡ ⎤ ⎡ ⎤−1 ⎡ ⎤ ⎡ ⎤ ⎡ I1 0.1250 0 0.3750 −0.1250 0.2500 5 1 −3 0 ⎣ I2 ⎦ = ⎣ 1 5 1 ⎦ ⎣ 1 ⎦ = ⎣ −0.1250 0.2500 −0.1250 ⎦ ⎣ 1 ⎦ = ⎣ 0.1250 ⎦ I3 0.2500 1 0.2500 −0.1250 0.3750 −3 1 5 1 {IR1 = 0.2500,

IR3 = 0.1250,

IR5 = 0.3750}.

2.3 Examples

33

Let us compute the power consumed by the circuit. The resistors are R1 = 1, R2 = 3, R3 = 3, R4 = 1, R5 = 1, R6 = 1. The currents through the resistors due to the source, respectively, are I1 = 0.25, I2 = 0.125, I3 = 0.125, I4 = 0.25, I5 = 0.375, I6 = 0.125. The power consumed is 0.375 W. The current supplied by the voltage source is 0.375. Therefore, the power supplied is 0.375 W, as obtained before. Circuit Analysis with a Supernode If a voltage source, independent or dependent, is not connected to the ground node, then its two nodes and any elements connected in parallel with it is called a supernode. The current through that source has to be found in order to solve the problem by nodal analysis. The voltage difference between the terminals of the source is the given constraint. Sufficient number of KVL and KCL equations has to be set up and solved to find the current through the voltage source between the two nonreference nodes. Consider the bridge circuit shown in Fig. 2.8. The voltage source is in the middle of the bridge. Same nodes and links are assumed as in earlier bridge circuit analysis. Usually, two equations using KCL is set up at the two ends of the supernode. Let the current through the source be i. At the left side node, we get (V2 − V1 ) V2 + = i. (2.1) 1 3 At the right side node, we get (V1 − (V2 − 1)) (1 − V2 ) + = i. 3 1

(2.2)

Eliminating i and simplifying, we get V1 − 2V2 = −1.

(2.3)

Applying KCL at node 1, we get (V1 − V2 ) (V1 − (V2 − 1)) V1 + + = 0. 1 1 3 Fig. 2.8 A bridge circuit with the voltage source in the middle

V1 I1 0.2A R1 R5

0.2V 0.4A 1Ω

I2 0.2A R2

3Ω

1V i 1Ω V2 V3 0.6A −0.4V 0.6V R3 3Ω R4 1Ω 0.2A I3 0.4A

34

2 DC Circuits

Note that the current in the source is flowing towards node 2. Since there is no other source in the circuit, this current gets split up at node 1. Simplifying, we get 7V1 − 4V2 = −1.

(2.4)

Solving Eqs. (2.3) and (2.4), we get V1 = 0.2

and

V2 = 0.6

V3 = V2 − 1 = 0.6 − 1 = −0.4 V. From Eq. (2.1), we get the current through the source as (0.6 − 0.2) 0.6 + = i = 0.6. 1 3 Since we have determined the two independent voltages, there is no need to use this current for the analysis of this circuit. {VR1 = −0.4, VR2 = 0.6}. An alternative method to avoid processing a supernode is that we connect a resistance in series with the source, as presented in an earlier example. As its value gets reduced compared with other resistances in the circuit, the circuit becomes more closer with a zero resistance series circuit and the result becomes closer to the exact values. A suitable small series resistance R is to be selected. For the example circuit with R = 0.1 , we get, by simulation, for V1 , V2 , V3 {0.1887, 0.566, −0.3774}. With R = 0.01 , we get With R = 0.001 , we get With R = 0.0001 , we get

{0.1988, 0.5964, −0.3976}. {0.1999, 0.5966, −0.3998}. {0.2, 0.6, −0.4}.

Mesh Analysis The loop current directions are assumed to be as in Fig. 2.2b. The equilibrium equations are similar to that in Fig. 2.6. Z1 (I1 − I2 ) + Z3 (I1 − I3 ) + Z5 I1 = 0 −Z1 (I1 − I2 ) + Z2 I2 = 1 −Z3 (I1 − I3 ) + Z4 I3 = −1. In matrix form, ⎤⎡ ⎤ ⎡ ⎤ I1 0 −Z1 −Z3 (Z1 + Z3 + Z5 ) ⎣ −Z1 (Z1 + Z2 ) 0 ⎦ ⎣ I2 ⎦ = ⎣ 1 ⎦ . −Z3 0 (Z3 + Z4 ) I3 −1 ⎡

2.3 Examples

35

With Z1 = 1,

Z2 = 3,

Z3 = 3,

Z4 = 1,

Z5 = 1,

V = 1,

⎤⎡ ⎤ ⎡ ⎤ I1 5 −1 −3 0 ⎣ −1 4 0 ⎦ ⎣ I2 ⎦ = ⎣ 1 ⎦ . I3 −3 0 4 −1 ⎡

we get

For this circuit, we can get the impedance matrix from inspection. The determinant of the impedance matrix is 40. ⎡

⎤ ⎤ ⎡ ⎤⎡ ⎤−1 ⎡ ⎤ ⎡ ⎤ ⎡ I1 −0.2 0 0.4000 0.1000 0.3000 5 −1 −3 0 ⎣ I2 ⎦ = ⎣ −1 4 0 ⎦ ⎣ 1 ⎦ = ⎣ 0.1000 0.2750 0.0750 ⎦ ⎣ 1 ⎦ = ⎣ 0.2 ⎦ . I3 −0.4 −1 0.3000 0.0750 0.4750 −3 0 4 −1 Figure 2.8 shows the circuit with correct current directions and the node voltages. Let us compute the power consumed by the circuit. The resistors are R1 = 1, R2 = 3, R3 = 3, R4 = 1, R5 = 1. The currents through the resistors due to the source, respectively, are I1 = 0.4, I2 = 0.2, I3 = 0.2, I4 = 0.4, I5 = 0.2. The power consumed is 0.6 W. The current supplied by the voltage source is 0.6. Therefore, the power supplied is 0.6 W, as obtained before. Current Sources Without sources, a circuit is dead with no currents through the elements and no voltage across the elements. In practice, there is no such thing as an ideal device. However, we analyze engineering systems with ideal elements first, since it simplifies the analysis. Then, the ideal element is modified to represent actual elements. An ideal current source is characterized by its volt–ampere relationship of keeping its terminal current same irrespective of the voltage across the load circuit connected to it, as shown in Fig. 2.9a. A practical current source can be approximated by an ideal current source with a shunt resistance in parallel with its terminals, as shown in Fig. 2.9b. Therefore, the current supplied by a practical current source will decrease as the terminal voltage is increased. A current source is an applied constraint in the circuit. Due to this constraint, the currents through various branches of the circuit are fixed depending on the characteristics of the current source and the circuit. An arrow indicates that a current raise in that direction is considered as a positive quantity. A voltage source appearing in series with a branch or a current source appearing in parallel with a branch does not Fig. 2.9 (a) Ideal and (b) practical current sources and their volt–ampere characteristics

It +

+ I

V

I

I

− (a)

V

R

V

It I

− (b)

V

36

2 DC Circuits

affect the structure of the circuit. That is, the number of independent voltages and currents remains the same. Two current sources with capacities I1 A and I2 A, connected in parallel, are equivalent to a single source with current capacity, which is equal to the algebraic sum of I1 and I2 . Two unequal current sources are not allowed to be connected in series, since it is a violation of KCL. Circuit Analysis with Current and Voltage Sources Consider a circuit energized by current and voltage sources, shown in Fig. 2.10a. One possible tree is shown in Fig. 2.10b. With two voltages known and the current constrained in one branch, both the nodal and loop analysis reduce to two variable problems. Let us do the nodal analysis. V1 = 1 is known. Therefore, we have to determine V2 and V3 only. The equilibrium equations at nodes 2 and 3, respectively, are (V2 − 1) V2 + = 1 or 4V2 = 6, V2 = 1.5 V 1 3 16 (V3 + 1) V3 + = −1 or 7V3 = −16, V3 = − V. 3 4 7 Mesh Analysis Applying KVL around the left side loop, we get, with I2 = −1, (I1 )1 + 3(I1 − I2 ) = 1 or 4I1 = −2, I1 = −0.5 A. Applying KVL around the right side loop, we get, with I2 = −1, 4 −(I2 − I3 )3 + 4I3 = −1 or 7I3 = −4, I3 = − A. 7 A supermesh occurs when a current source is connected between two nonreference nodes without a parallel resistance. We apply KVL to the neighboring loops to find the voltage across the current source. An alternative method to avoid processing a supermesh is that we connect a resistance in parallel with the current source. As the value gets increased compared with other resistances in the circuit, the circuit becomes more closer with an infinite resistance parallel circuit and the result

V2 I1

1A V3

R3 2Ω

V3

I3 R4

R1 1Ω V1

V2

I2

3Ω R5 4Ω

R2 3Ω

R1

R4

1Ω

3Ω

V1 1V

1V

1V

1V (a)

(b)

Fig. 2.10 (a) A circuit energized by current and voltage sources and (b) a possible tree

2.3 Examples

37

becomes closer to the exact values. A suitable high parallel resistance R is to be selected. For the example circuit, with R = 1000 , we get, by simulation, {I1 = −0.496 A,

I2 = −0.9942 A,

I3 = −0.569 A}.

With R = 10000 , we get {−0.5 A,

2.3.1

−0.9994 A,

−0.5713 A}.

Linearity Property of Circuits

This property is called the superposition theorem. When circuits consist of linear elements, such as resistors, inductors, capacitors or, equivalently, circuits characterized by a linear differential equation with constant coefficients, the circuits are linear circuits. Most of the systems, although to some extent are nonlinear, can be approximated adequately by linear systems. For example, in theoretical analysis, we assume a pure resistor with no other parasitic elements. A resistor may be nonlinear to some extent and may have some inductive effect. It is assumed that the resistor is sufficiently pure for the accuracies required in practical applications. As the assumed linearity provides a simpler analysis, linear systems analysis is widely used in the study of signals and systems. Similarly, signals, usually, have arbitrary amplitude profile and are decomposed into a linear combination of well-defined basic signals in transform analysis, as presented later. Mathematically, the response of a circuit to a linear combination of inputs is linear if the output is also the same linear combination of the individual outputs to the respective inputs. For example, let the response of a circuit to 1 V input is 1 A. Then, the output must be 2 A for the input 2 V. Therefore, the analysis becomes somewhat simpler with a normalized input voltage of 1 V, although the electric power is supplied at 220/110 V. And it is much higher in power transmission systems. Further, typical resistor values are in kiloohms or megaohms. These aspects are to be learnt in a concurrent laboratory course. Another implication of linearity is that the output of a circuit, with a number of sources, can be obtained by finding the output to each source separately and adding all the responses. Voltages and currents due to each source and their total in the circuit, shown in Fig. 2.10, are shown in Table 2.1. Determining the response to each source is much simpler. The use of this property may result in simpler analysis, compared with nodal and loop methods. The circuit has two voltage sources and one current source. When the response to one source is considered, the rest of the voltage sources are replaced by short-circuits and the current sources are replaced by open-circuits. Let us consider the response to the left voltage source alone, case (a). The right voltage source is short-circuited and the current source is open-circuited. The response is shown in the left side of Fig. 2.11a. Let us consider the response to the right voltage source alone, case (b). The left voltage source is short-circuited and the current source is open-circuited. The response is shown in the right side of Fig. 2.11a. Let us consider the response to the current source alone. The two voltage sources are short-circuited. The Table 2.1 Voltages and currents due to each source and their total in the circuit shown in Fig. 2.10

Case

V1

V2

V3

I1

I2

I3

(a) (b) I2 only Total

1 0 0 1

0.75 0 0.75 1.5

0 − 74 − 12 7 − 16 7

0.25 0 −0.75 −0.5

0 0 −1 −1

0 − 71 − 73 − 74

38

2 DC Circuits

V3 = − 74 V I3

V2 = 0.75V I1 0.25A R1 V1

R4

1Ω

3Ω R5 1 7A

R2 3Ω 1V

V2 = 0.75V R3 I1 0.75A 2Ω 0.25A 4Ω R1 1Ω 1 7A

1A V3 = − 12 V 7

I2

I3 R4

R2 3Ω

3Ω R5 4Ω 4 7A

3 7A

1V (a)

(b) (a)

(b)

Fig. 2.11 Analysis of a circuit using the linearity property

response is shown in Fig. 2.11b. The currents are obtained by using the current-division formula for resistors connected in parallel. Let us compute the power consumed by the circuit. The resistors are R1 = 1, R2 = 3, R3 = 2, R4 = 3, R5 = 4. The net currents through the resistors due to all the sources, respectively, are I1 = 0.5, I2 = 0.5, I3 = 1, I4 =

3 4 , I5 = . 7 7

The power consumed is 27 64 + = 4.8571 W. 49 49 The equation defining the power is nonlinear. Therefore, the total power supplied may not be equal to the sum of the powers supplied by the sources acting alone. Let the currents supplied by two sources acting alone be I1 and I2 . Then, the sum of the powers supplied is proportional to I12 + I22 . The power supplied by the sources acting together is proportional to (I1 + I2 )2 and 0.25 + 0.75 + 2 +

(I12 + I22 ) = (I1 + I2 )2 . The total power is always equal to the sum of the powers consumed by the separate branches of the circuit, irrespective of the number of sources in the circuit. When all the sources acting simultaneously, the total power supplied by the sources is the sum of the powers supplied by the individual sources. The power supplied by a source is the product of its terminal voltage and current. The voltage across a current source depends jointly upon the intensities of all the sources and vice versa for a voltage source. Therefore, the power supplied by a source can vary, when it is acting alone or jointly with other sources. With all the sources acting, the net current through the left voltage source is −0.5. Therefore, it consumes power, which is equal to −0.5 W. Similarly, the right voltage source consumes −3/7 W. The voltage across the current source is 1.5 + 2 + 16/7, when it is acting jointly. Therefore, the total power is −0.5 − 3/7 + (1.5 + 2 + 16/7) = 4.8571 W, as obtained before.

2.3 Examples

39

R1 I1

3Ω

V1

V2

V2

I2 R2 1Ω

1V

I3 R3

I3 V1

4Ω

R3

1V

2A (a)

4Ω

(b)

Fig. 2.12 (a) A circuit with a current source between two loops and (b) a possible tree

Circuit Analysis with the Current Source Between Two Loops Consider the circuit with a current source between two loops, shown in Fig. 2.12a. A possible tree is shown in Fig. 2.12b. With 3 nodes and one independent voltage and one independent current given, the analysis reduces to a one-variable problem. Mesh Analysis With I1 − I3 = 2 A and applying KVL, we get 3I1 + 4I3 = 1 I1 − I3 = 2. Solving, we get

9 I1 = = 1.2857, 7

5 I3 = − = −0.7143 7

and V2 = −20/7 V. Current in the rightmost resistor flows opposite to the direction of initial assumption. Nodal Analysis V2 − 1 20 V2 + = −2 and V2 = − V. 4 3 7 Let us compute the power consumed by the circuit. The resistors are R1 = 3, R2 = 1, R3 = 4. The net currents through the resistors due to all the sources, respectively, are I1 = 1.2857, I2 = 2, I3 = 0.7143. The power consumed is 1.28572 (3) + 22 + 0.71432 (4) = 11 W. The net current through the voltage source is 9/7. The voltage across the current source is (−20/7−2), when it acts jointly. Therefore, the total power is 9/7 + (−2)(−34/7) = 11 W as obtained before.

40

2 DC Circuits

Fig. 2.13 Voltage-controlled voltage source and its volt–ampere characteristics

I1 = 0 +

I2 +

V1

kV1 + − V2

−

−

V2

(b) I2

(a) Fig. 2.14 Current-controlled current source and its volt–ampere characteristics

I1 + V1 = 0

I2 + kI1

−

V2 = kV1

V2

I2

I2 = kI1

− (a)

(b) V2

Controlled Sources Sources can both supply and absorb power. In charging, a battery absorbs power. It delivers power when connected to a device, such as a light bulb. Batteries and generators are commonly used to energize circuits and those are approximations to ideal sources. The voltage or current provided by a controlled or dependent source, a 3-terminal device, is controlled by another voltage or current at some other part of the circuit. The 3 terminals, with one common terminal, form two pairs, one pair for input and the other for the output. Four possible types are as follows: 1. 2. 3. 4.

voltage-controlled voltage source current-controlled voltage source voltage-controlled current source current-controlled current source

Dependent sources are necessary to model devices such as transistors. Voltage-controlled voltage source and its volt–ampere characteristics are shown in Fig. 2.13. It is a 3-terminal device. The source voltage is k, a constant, times the control voltage V1 , irrespective of the value of I2 . Current-controlled current source and its volt–ampere characteristics are shown in Fig. 2.14. The source current is k, a constant, times the control current I1 , irrespective of the value of V2 .

2.3.2

Analysis of a Circuit with a Controlled Voltage Source

A circuit with a voltage-controlled voltage source is shown in Fig. 2.15. A possible tree corresponding to the circuit is shown in Fig. 2.16. Loops corresponding to the currents I1 , I2 , and I3 are shown in Figs. 2.17 and 2.18. The voltage-controlled voltage source generates a voltage of 2(V1 − V4 ). The circuit has five nodes. As there is a dependent source and a branch current is known, the analysis involves only three independent current or voltage variables. Let us analyze the circuit using the loop method. Every element has to be included in a loop. At least one element in each loop is not part of any other independent loop. The equilibrium equations corresponding to the middle, left, and right loops, respectively, are

2.3 Examples

41

Fig. 2.15 A circuit with a voltage-controlled voltage source

R5 I1 −4.75A + −

V1 −3.5V

2Ω V2 ,15.5V R2 V3 ,7V 1V

−1.75A 2(V1 − V4 ) R1 2Ω

I2 1.75A

1Ω

R3

4Ω

V4 6V I3 2A R4

3Ω

2A

Fig. 2.16 A possible tree corresponding to the circuit in Fig. 2.15

V1

R2

V2

V3 1V

V4

1Ω R1 2Ω 2A

Fig. 2.17 The loop corresponding to current I1

R5 I1 V2

V1

2Ω R2

V3 1V

V4

1Ω R1 2Ω 2A

Fig. 2.18 Loops corresponding to the currents I2 and I3

V2

R2 1Ω

V3

V2

I2 R3

2A

R2

V3 1V

1Ω

V4 I3 R4

4Ω

3Ω

2A

(Z5 I1 + Z4 I3 − 2Z5 I1 ) + (Z2 (I1 − I2 − I3 )) − Z3 I2 = 0 −Z1 (I (2) + I (3) − 2) = Z5 I1 + Z4 I3 Z3 I2 − Z4 I3 = 1. The first equation corresponds to the middle loop involving resistors Z2 and Z3 . To write the KVL equation around the loop, we need the voltage V2 . V1 = Z5 I1 + Z4 I3

and

V1 − V4 = Z5 I1 .

42

2 DC Circuits

Therefore, the voltage across the controlled voltage source is 2Z5 I1 , with opposite polarity to that of V1 . Consequently, V2 = (Z5 I1 + Z4 I3 − 2Z5 I1 ). The current IZ2 is (I1 − I2 − I3 ). The second equation corresponds to the leftmost loop involving resistor Z1 . We have the equation V1 = Z5 I1 + Z4 I3 . The current entering Z1 is (I (2)+I (3)−2) from the ground side. Therefore, we get another expression for V1 involving the current source. V1 = −Z1 (I (2) + I (3) − 2). Equating these two equations yields an equilibrium equation. The last equation corresponds to the rightmost loop involving resistors Z3 and Z4 . Simplifying, the three equilibrium equations are (−Z5 + Z2 )I1 − (Z2 + Z3 )I2 + (Z4 − Z2 )I3 = 0 Z5 I1 + Z1 I2 + (Z1 + Z4 )I3 = 4 0I1 + Z3 I2 − Z4 I3 = 1. With Z1 = 2, we get

Z2 = 1,

Z3 = 4,

Z4 = 3,

Z5 = 2,

⎤⎡ ⎤ ⎡ ⎤ 0 −1 −5 2 I1 ⎣ 2 2 5 ⎦ ⎣ I2 ⎦ = ⎣ 4 ⎦ . I3 1 0 4 −3 ⎡

The determinant of the impedance matrix is 12. ⎡

⎤ ⎡ ⎤−1 ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ I1 −1 −5 2 0 −2.1667 −0.5833 −2.4167 0 −4.75 ⎣ I2 ⎦ = ⎣ 2 2 5 ⎦ ⎣ 4 ⎦ = ⎣ 0.5000 0.2500 0.7500 ⎦ ⎣ 4 ⎦ = ⎣ 1.75 ⎦ . I3 0 4 −3 1 0.6667 0.3333 0.6667 1 2 The other currents and voltages are shown in Fig. 2.15. Nodal Analysis The voltage source is not connected to the ground node. Therefore, we have to find the current through it. Setting up two KCL equations at the terminals of the source, we get V4 (V3 − V2 ) V3 (V4 − V1 ) + + + = 0. Z5 Z4 Z2 Z3 The sum of the first two terms is the current leaving from node 4 towards node 3, which is the negative of the current leaving from node 3 towards node 4. The dependent voltage source is also not connected to the ground node. Therefore, we have to find the current through it. Setting up two KCL equations at the terminals of the source, we get

2.3 Examples

43

−

(V2 − V3 ) (V1 − V4 ) V1 + (2) − − = 0. Z2 Z5 Z1

Further, V1 − V2 = 2(V1 − V4 )

or

V1 + V2 − 2V4 = 0.

The equilibrium equations are −

(V2 − V3 ) (V1 − V4 ) V1 + (2) − − =0 Z2 Z5 Z1

(2.5)

(V4 − V1 ) V4 (V3 − V2 ) V3 + + + =0 Z5 Z4 Z2 Z3

(2.6)

V1 + V2 − 2V4 = 0.

(2.7)

Replacing

V3 = (V4 + 1),

we get ((V4 + 1) − V2 ) (V1 − V4 ) V1 + (2) − − =0 Z2 Z5 Z1

(2.8)

(V4 − V1 ) V4 (V2 − (V4 + 1)) (V4 + 1) + − + =0 Z5 Z4 Z2 Z3

(2.9)

V1 + V2 − 2V4 = 0.

(2.10)

Simplifying, we get −(Z1 Z2 + Z2 Z5 )V1 − Z1 Z5 V2 + (Z1 Z2 + Z1 Z5 )V4 = −Z1 Z5 − 2Z1 Z2 Z5 −Z2 Z3 Z4 V1 − Z3 Z4 Z5 V2 + (Z2 Z3 Z4 +Z2 Z3 Z5 + Z3 Z4 Z5 + Z2 Z4 Z5 )V4 = −Z2 Z4 Z5 − Z3 Z4 Z5 V1 + V2 − 2V4 = 0. With Z1 = 2, we get

Z2 = 1,

Z3 = 4,

Z4 = 3,

Z5 = 2,

⎤ ⎤⎡ ⎤ ⎡ −12 V1 −4 −4 6 ⎣ −12 −24 50 ⎦ ⎣ V2 ⎦ = ⎣ −30 ⎦ . V4 0 1 1 −2 ⎡

The determinant of the admittance matrix is −24. ⎡

⎤−1 ⎡ ⎤ ⎤ ⎡ V1 −4 −4 6 −12 ⎣ V2 ⎦ = ⎣ −12 −24 50 ⎦ ⎣ −30 ⎦ V4 1 1 −2 0 ⎤ ⎤ ⎡ ⎤⎡ ⎡ −3.5 −12 0.0833 0.0833 2.3333 = ⎣ −1.0833 −0.0833 −5.3333 ⎦ ⎣ −30 ⎦ = ⎣ 15.5 ⎦ . 6 0 −0.5000 0 −2.0000 The voltages are the same as those found using loop analysis.

44

2 DC Circuits

The resistors and the respective currents are R1 = 2, R2 = 1, R3 = 4, R4 = 3, R5 = 2 and I 1 = 1.75, I 2 = 8.5, I 3 = 1.75, I 4 = 2, I 5 = 4.75. The power consumed is 147.75 W. P = I 12 R1 + I 22 R2 + I 32 R3 + I 42 R4 + I 52 R5 = 147.75 W The power supplied is (−6.5)(−19) − (1)(6.75) + 2(15.5) = 147.75, as obtained above.

2.3.3

Analysis of a Circuit with a Controlled Current Source

Nodal Analysis A circuit with a current-controlled current source is shown in Fig. 2.19. A tree corresponding to the circuit is shown in Fig. 2.20. Since V3 = 1 V is known, there are only 2 unknowns, {V1 , V2 }. The two equilibrium equations are (V2 − V1 ) (V2 − 1) V2 + + =0 Z2 Z4 Z3 Fig. 2.19 A circuit with a current-controlled current source

R5 I1 0.4167 V1 1.833V Ix 0.9167

2Ω R2 V2 4 3 1 1Ω I2 3

R1 2Ω

R3

R4 2Ω

4Ω

V3 1 I3 0.5833 1V

2Ix 1.833A

Fig. 2.20 A possible tree corresponding to the circuit in Fig. 2.19

V1 Ix R1 2 Ω

V2

V3

2.3 Examples

45

V1 (V1 − 1) V1 (V1 − V2 ) + −2 + = 0. Z1 Z5 Z1 Z2 The first equation is due to the application of KCL at node with voltage V2 . The second equation is due to the application of KCL at node with voltage V1 . Simplifying, we get −Z3 Z4 V1 + (Z3 Z4 + Z2 Z4 + Z2 Z3 )V2 = Z2 Z3 (−Z2 Z5 + Z1 Z5 + Z1 Z2 )V1 − Z1 Z5 V2 = Z1 Z2 . With Z1 = 2,

Z2 = 1,

Z3 = 4,

Z4 = 2,

Z5 = 2,

we get −4V1 + 7V2 = 2 2V1 − 2V2 = 1. Solving for the voltages, we get



V1 V2

 =

11  6 4 3

.

All the currents and voltages are shown in Fig. 2.19.

Mesh Analysis The three loops are shown in Fig. 2.21. In a controlled source, the output current does not have any influence on the input. Therefore, IZ1 = I2 + I3 . The equilibrium equations are Z1 (I2 + I3 ) + Z2 (I1 − I2 − I3 ) − Z3 I2 = 0 Z3 I2 + Z4 (I1 − I3 ) = 1 Z1 (I2 + I3 ) − Z5 I1 = 1. R5 I1 V1

2Ω R2

V2

1Ω V1

R2

Ix

1Ω

R1 2Ω

V2 I2 R3

R4

V3

2Ω

V1

R2

Ix

1Ω

4Ω R1 2Ω

Fig. 2.21 A circuit with a controlled current source, the three loops

V2

R4 2Ω

V3 I3 1V

46

2 DC Circuits

The first equation is obtained by equating the voltages on both sides of the current source. The second equation is obtained by applying KVL to the rightmost loop involving the voltage source. The third equation is obtained by applying KVL to the loop involving the voltage source and impedances Z1 and Z5 . Voltage source is 1 V. Simplifying, we get Z2 I1 + (Z1 − Z2 − Z3 )I2 + (Z1 − Z2 )I3 = 0 Z4 I1 + Z3 I2 − Z4 I3 = 1 −Z5 I1 + Z1 I2 + Z1 I3 = 1. With Z1 = 2, we get

Z2 = 1,

Z3 = 4,

Z4 = 2,

Z5 = 2,

⎡

⎤⎡ ⎤ ⎡ ⎤ 1 −3 1 I1 0 ⎣ 2 4 −2 ⎦ ⎣ I2 ⎦ = ⎣ 1 ⎦ . I3 −2 2 2 1

The determinant of the impedance matrix is nonzero, 24. ⎤ ⎡ 1 −3 I1 ⎣ I2 ⎦ = ⎣ 2 4 I3 −2 2 ⎡ 0.5000 =⎣ 0 0.5000 ⎡

⎤−1 ⎡ ⎤ 1 0 −2 ⎦ ⎣ 1 ⎦ 2 1 ⎤ ⎤⎡ ⎤ ⎡ 0.4167 0 0.3333 0.0833 0.1667 0.1667 ⎦ ⎣ 1 ⎦ = ⎣ 0.3333 ⎦ . 0.5833 1 0.1667 0.4167

The resistors and the respective currents are R1 = 2, R2 = 1, R3 = 4, R4 = 2, R5 = 2 and I 1 = 0.9167, I 2 = −0.4997, I 3 = 0.3333, I 4 = −0.1666, I 5 = 0.4167. The power consumed is 2.7776 W. P = I 12 R1 + I 22 R2 + I 32 R3 + I 42 R4 + I 52 R5 = 2.7777 W The power supplied by the sources is −0.5833 + (11/6)2 = 2.7777.

Circuits with One or Two Variables Now, we are going to present the analysis of simpler circuits with one or two variables, which can be solved manually. Circuit in Fig. 2.22 The circuit shown in Fig. 2.22 has two voltage sources. Nodal Analysis Z1 = 2,

Z2 = 4,

Z3 = 3,

V1 = 1,

V3 = 2

2.3 Examples

47

Fig. 2.22 A circuit with two voltage sources

R1 I1

2Ω

V1

V2 I2 R2

1V

R3 3Ω I3 V3 4Ω 2V

Voltage V2 is the only unknown. Therefore, applying KCL at node 2, we get (V2 − 1) (V2 − 2) V2 + + = 0. Z1 Z3 Z2 Solving for V2 , we get V2 =

14 Z2 Z3 + 2Z1 Z2 28 = V. = Z1 Z2 + Z2 Z3 + Z1 Z3 26 13

The currents are I1 = −

(V2 − 1) 1 = − A, Z1 26

I2 =

7 (V2 ) 14 = A, = Z2 52 26

I3 =

(V2 − 2) 4 = − A. Z3 13

With −I1 + I2 + I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 390 15 1 = 0.5769 W. (2 + 49 × 4 + 64 × 3) = 2 = 2 26 26 26 The total power supplied by the sources is −

16 15 1 + = = 0.5769 W. 26 26 26

Current enters the positive terminal of the 1 V source, while current leaves in the 2 V source. Therefore, 1 V source absorbs power, while 2 V source delivers power. If the current flow in a voltage source is into the positive terminal, then it absorbs power and the power is negative. If the current flow in a voltage source is out of the positive terminal, then it delivers power and the power is positive. While the current from an ideal current source is fixed, the voltage across it may be positive or negative, depending on the circuit constraints. If the current flow in a current source is into the positive voltage terminal, then it absorbs power and the power is negative. If the current flow in a current source is out of the positive voltage terminal, then it delivers power and the power is positive. Mesh Analysis I2 = I1 − I3 . Z1 I1 + Z2 (I1 − I3 ) = 1 Z3 I3 − Z2 (I1 − I3 ) = −2.

48

2 DC Circuits

Simplifying, we get (Z1 + Z2 )I1 − Z2 I3 = 1 −Z2 I1 + (Z2 + Z3 )I3 = −2. Substituting the numerical values, we get 6I1 − 4I3 = 1 −4I1 + 7I3 = −2. Solving the equations, we get the same values of currents, as found earlier. Let us do the problem by superposition method. Let us find the response to 1 V source. Then, we replace 2 V source by a short-circuit. Current I1 is I1 =

1 7 = . Z1 + (Z2 ||Z3 ) 26

By current division, we get I2 = I1

Z3 3 4 , I3 = I1 − I2 = . = Z2 + Z3 26 26

Let us find the response to 2 V source. Then, we replace 1 V source by a short-circuit. Current I3 is I3 = −

6 2 =− . Z3 + (Z2 ||Z1 ) 13

By current division, we get I1 = I3

Z2 4 2 . = − , I2 = I1 − I3 = Z2 + Z1 13 13

Adding the corresponding two responses, we get the same values for the currents.

Circuit in Fig. 2.23 The circuit shown in Fig. 2.23 has voltage and current sources. Fig. 2.23 A circuit with voltage and current sources

R1 I1

2Ω

V2 I2 R2

1A

R3 3Ω I3 V3 4Ω 2V

2.3 Examples

49

Nodal Analysis Z1 = 2,

Z2 = 4,

Z3 = 3,

I1 = −1,

V3 = 2.

Voltage V2 is the only unknown. Therefore, applying KCL at node 2, we get (V2 − 2) V2 + = −1. Z3 Z2 Solving for V2 , we get V2 =

−Z2 Z3 + 2Z2 4 = − V. Z2 + Z3 7

The currents are I1 = −1 A,

I2 =

V2 1 = − A, Z2 7

I3 =

(V2 − 2) 6 = − A. Z3 7

With I1 − I2 − I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 2+

112 30 1 = 4.2857 W. (1 × 4 + 36 × 3) = 2 + 2 = 7 72 7

The total power supplied by the sources is   4 30 6 2+ = 4.2857 W. +2 = 7 7 7 Both sources deliver power. Mesh Analysis I2 = I1 − I3 , I1 = −1. Z3 I3 − Z2 (I1 − I3 ) = −2. Solving for I3 , we get (Z3 + Z2 )I3 = −6,

6 I3 = − A. 7

We get the same values of currents, as found earlier. Let us solve the problem by superposition method. Let us find the response due to voltage source. The current source is open-circuited. Then, I3 = −2/7, I2 = 2/7. Let us find the response due to current source. The voltage source is short-circuited. Then, by current division, I3 = −4/7, I2 = −3/7. Adding the corresponding two responses, we get the same values for the currents.

Circuit in Fig. 2.24 The circuit shown in Fig. 2.24 has voltage and current sources. Source voltage is 2 V. Source current is 1 A. Z1 = 2, Z2 = 4, Z3 = 3, I3 = 1.

50

2 DC Circuits

Fig. 2.24 A circuit with voltage and current sources

R1 I1

2Ω

R3

V2 I2 R2

3Ω I3 4Ω 1A

2V

Mesh Analysis I1 − I3 = I2 I1 Z1 + I2 Z2 = I1 Z1 + (I1 − I3 )Z2 = 2,

I1 =

2 + I3 Z2 = 1, Z2 + Z1

I2 = 0.

With −I1 + I2 + I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 3 + 2 = 5 W. The total power supplied by the sources is 3+2=

30 = 5 W. 7

Both sources deliver power. Nodal Analysis V2 (V2 − 2) + = −1. Z2 Z1 Solving, V2 = 0 and VR3 = 3 V. The voltage across the current source is −3 V. Let us do the problem by superposition method. Consider the response due to the voltage source alone. The current source is open-circuited. Then, I1 = I2 =

1 . 3

Consider the response due to the current source alone. The voltage source is short-circuited. Then, by current division, we get 2 1 I1 = , I2 = − . 3 3 Adding the partial currents, we get the same results.

Circuit in Fig. 2.25 The circuit shown in Fig. 2.25 has current sources. Source currents 2 and 1 A. Z1 = 2,

Z2 = 4,

Z3 = 3.

2.3 Examples

51

Fig. 2.25 A circuit with current sources

R1 I1

2Ω

R3

V2 I2 R2

3Ω I3 4Ω

2A

1A

Nodal Analysis V2 = −3 and V2 = −12 V Z2 I1 = −2, I2 = −

12 = −3 A, I3 = 1 A 4

With −I1 + I2 + I3 = 0, KCL is satisfied. Mesh Analysis I3 = 1,

I1 = −2,

I2 = (I1 − I3 ) = −3 A

Let us do the problem by superposition method. Consider the response due to the 2 A current source alone. The other current source is open-circuited. Then, we get I1 = −2,

I2 = −2,

I3 = 0

Consider the response due to the 1 A current source alone. The other current source is open-circuited. Then, by current division, we get I1 = 0,

I2 = −1,

I3 = 1.

Adding the two partial results, we get the same currents. The total power dissipated by the resistors is 8 + 36 + 3 = 47 W. The total power supplied by the sources is 16 × 2 + 15 = 47 W. Both sources deliver power.

Circuit in Fig. 2.26 A circuit with a voltage-controlled voltage source is shown in Fig. 2.26. Source voltage is 1 V. Z1 = 2,

Z2 = 4,

Z3 = 3.

52

2 DC Circuits

Fig. 2.26 A circuit with a voltage-controlled voltage source

V2 I1 R1 V1

I2 2Ω

R3 I3 3Ω + − (V2 −V1 )

R2 4Ω 1V

Nodal Analysis (V2 − 1) V2 V2 − (V2 − 1) + + = 0. 2 4 3 Solving, we get V2 = 2/9. The currents are I1 = −

7 A, 18

I2 = −(I3 + I1 ) =

1 A, 18

I3 =

1 A. 3

With I1 + I2 + I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 35 1 = 0.6481 W. (98 + 4 + 108) = 2 54 18 The total power supplied by the sources is 7 (−7) (−1) 35 + = = 0.6481 W. 18 9 3 54 Both sources deliver power. Mesh Analysis 2I1 + 4(I3 + I1 ) = −1, 3I3 + 2I1 + 4(I3 + I1 ) = 0,

or or

6I1 + 4I3 = −1 6I1 + 7I3 = 0.

Solving, we get the same currents obtained earlier.

Circuit in Fig. 2.27 A circuit with a current-controlled voltage source is shown in Fig. 2.27. Source voltage is 1 V. Z1 = 2,

Z2 = 4,

Z3 = 3.

2.3 Examples

53

Fig. 2.27 A circuit with a current-controlled voltage source

I1 R1 V1

V2

R3

I2

3Ω

2Ω R2 4Ω 1V

Nodal Analysis Since 4I2 = V2 , 2I2 = 0.5V2 . V2 − 0.5V2 (V2 − 1) V2 + + = 0. 2 4 3 Solving, we get V2 = 6/11. The currents are I1 = −

5 A, 22

I2 =

3 A, 22

I3 =

1 A. 11

With I1 + I2 + I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 98 1 (50 + 36 + 12) = 2 = 0.2025 W. 2 22 22 The total power supplied by the sources is 5 3 (−1) + = 0.2025 W. 22 11 11 The controlled source absorbs power. Mesh Analysis

2I1 + 4(I3 + I1 ) = −1, 3I3 − 2(I3 + I1 ) + 4(I3 + I1 ) = 0,

or or

6I1 + 4I3 = −1 2I1 + 5I3 = 0.

Solving, we get the same currents obtained earlier.

Circuit in Fig. 2.28 A circuit with a voltage-controlled current source is shown in Fig. 2.28. Source current is 1 A. Z1 = 2,

Z2 = 4,

Z3 = 3.

I3 + − 2I2

54

2 DC Circuits

Fig. 2.28 A circuit with a voltage-controlled current source

I1 R1

V2

R3

I2

3Ω

I3 2V2

2Ω R2 4Ω 1A

Nodal Analysis −1 − 2V2 +

V2 = 0. 4

Solving, we get V2 = −4/7. The currents are I1 = −1 A,

1 I2 = − A, 7

I3 =

8 A. 7

With I1 + I2 + I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 2+

192 4 + = 6 W. 49 49

The total power supplied by the sources is 10 (−8) 42 (1) + (−4) = = 6 W. 7 7 7 Both sources deliver power. Mesh Analysis 4(2V2 + 1) = V2 ,

V2 = −4/7 V

or

as before.

Circuit in Fig. 2.29 A circuit with a current-controlled current source is shown in Fig. 2.29. Source current is 1 A. Z1 = 2,

Z2 = 4,

Z3 = 3.

Nodal Analysis −1 − 0.5V2 +

V2 = 0. 4

2.3 Examples

55

Fig. 2.29 A circuit with a current-controlled current source

I1 R1

V2

R3

I2

3Ω

I3 2I2

2Ω R2 4Ω 1A

Solving, we get V2 = −4. The currents are I1 = −1 A,

I2 = −1 A,

I3 = 2 A.

With I1 + I2 + I3 = 0, KCL is satisfied. The total power dissipated by the resistors is 2 + 4 + 12 = 18 W. The total power supplied by the sources is (−2)(1) + (−2)(−10) = 18 W. Current source 1 A absorbs power. Mesh Analysis Since 4I2 = V2 , 2I2 = 0.5V2 . 4(0.5V2 + 1) = V2 ,

or

V2 = −4 V

as before.

2.3.4

Y −  and  − Y Transformations

In major applications, such as three-phase power systems and electrical filters, certain configurations of circuits often appear and it is required to transform one form from another for simpler circuit analysis. One is called the Y (wye) circuit, since it can be drawn resembling the letter Y , shown in Fig. 2.30a. That is, it has a common point between its three branches. The other one is called the  (delta) circuit, since it can be drawn resembling the Greek letter , shown in Fig. 2.30b. That is, it has no common point and its three branches are circularly connected in series. They are also called star-delta and T-π configurations. The transformation is such that the impedance between each pair of terminals remains the same before and after transformation.  − Y Transformation The impedance between terminals a and b in the Y circuit is Za + Zb . The impedance between terminals a and b in the  circuit is the parallel connection of Zab and Zac + Zbc . Since the values of the impedances between the terminals have to be same in both the configurations, we get

56

2 DC Circuits

Fig. 2.30 Y −  circuits

Za

a

a Zab

n

Zb

b

Zac

b Zbc

Zc (a)

Za + Zb =

Zab (Zac + Zbc ) . Zac + Zab + Zbc

Za + Zc =

Zac (Zab + Zbc ) Zac + Zab + Zbc

Zb + Zc =

Zbc (Zab + Zac ) . Zac + Zab + Zbc

c

c (b)

Similarly, we get

Solving the three equations, we get Za =

Zab Zac Zac + Zab + Zbc

Zb =

Zba Zbc Zac + Zab + Zbc

Zc =

Zca Zcb . Zac + Zab + Zbc

In words, the Y -impedances are given by the product of the two -impedances connected at the corresponding node divided by the sum of the three -impedances. We can solve the first set of equations to get the Y −  transformations. However, it is simpler to get this using duality property.

Duality A number and its logarithm are equivalent representation of the same entity. However, the logarithmic representation reduces the difficult multiplication problem into the simpler addition problem. We use the more useful representation of the two to solve a problem. Similarly, transform methods reduce the problem of solving a differential equation into solving algebraic equations. The formulation of circuit analysis problem using loop analysis is of the form IZ = V, where I is the current variable, V is the voltage variable, and Z is the resistance in the case of DC circuits. Let us replace I by V , V by I , and Z by 1/Z. Then, the equation becomes V = I = V G, Z

2.3 Examples

57

Table 2.2 Dual concepts and variables

Voltage

Current

Loop Resistance Inductance Voltage source KVL Short-circuit Parallel paths

Node Conductance Capacitance Current source KCL Open-circuit Series paths

where G is the conductance if Z is the resistance. This equation is another form of formulation of the circuit analysis problem using nodal analysis. The roles played by the voltage and current variables are interchanged. Since the variables play a dual role, they are called as dual variables. We know the nodal analysis is simpler for some circuits and vice versa. Obviously, we use the simpler one. Voltage is the dual of current and conductance is the dual of resistance. Consider again the finding of the equivalent resistor Req of connecting resistors in series and parallel. If the resistors R1 and R2 are connected in series, we get Req = R1 + R2 . If the resistors are connected in parallel, we get 1 1 1 = + Req R1 R2

or

Req =

R1 R 2 1 = . R1 + R2 G1 + G2

It is obvious that use of conductance is more convenient when analyzing parallel circuits. Some of the dual concepts and variables are shown in Table 2.2. Y −  Transformation Using the duality and assuming that the circuit consists of resistors, we get 1 = Zbc 1 = Zac 1 = Zab

1 Za

1 Za

1 Za

1 Zb

×

1 Zc

+

1 Zb

+

1 Za

×

1 Zc

+

1 Zb

+

1 Za

×

1 Zb

+

1 Zb

+

1 Zc

1 Zc

1 Zc

= Gbc =

Gb Gc Ga + Gb + Gc

or

Zbc =

Za Zb + Zb Zc + Zc Za Za

= Gac =

Ga Gc Ga + Gb + Gc

or

Zac =

Za Zb + Zb Zc + Zc Za Zb

= Gab =

Ga Gb Ga + Gb + Gc

or

Zab =

Za Zb + Zb Zc + Zc Za . Zc

In words, the -impedance connecting nodes a and b is obtained by dividing the sum of the pairwise products of the three Y -impedances by the Y -impedance at the opposite node, Zc . Similar interpretations hold for the other two impedances. If all the impedances are equal, it is called a balanced circuit. Then, Z and Z = 3ZY . ZY = 3 The difference in the conversion formulas for balanced circuits is due to the fact that Y -connection is similar to a series circuit while the other is similar to a parallel circuit.

58

2 DC Circuits

Fig. 2.31 A bridge circuit

I = 0.6A

c 0.6Ω

R1

1Ω

R2

3Ω

R5

1V

1V

1Ω R3

3Ω

R4

1Ω

0.2Ω a R3

(a)

3Ω

0.6Ω b R4

1Ω

(b)

Example Find the current I drawn by the bridge circuit, shown in Fig. 2.31a, from the source. The circuit can be considered as the combination of two  circuits with a shared resistor. To make the analysis simpler, we have to convert one of them into an equivalent circuit. Replacing the top half of the  circuit, we get, with Zbc = 3, Zac = 1, Zab = 1 1×1 = 0.2 3+1+1 3×1 Zb = = 0.6 3+1+1 3×1 Zc = = 0.6. 3+1+1

Za =

The transformed circuit is shown in Fig. 2.31b. Now, Zeq = 0.6 +

5 (0.6 + 1)(3 + 0.2) = . (0.6 + 1 + 3 + 0.2) 3

Therefore, I=

V 3 = (1) = 0.6 A, Zeq 5

which is the same as found by nodal and loop analyses. Let us get back the  circuit from the Y circuit. Zbc =

Za Zb + Zb Zc + Zc Za (0.2)(0.6) + (0.2)(0.6) + (0.6)(0.6) =3 = Za 0.2

Zac =

Za Zb + Zb Zc + Zc Za (0.2)(0.6) + (0.2)(0.6) + (0.6)(0.6) =1 = Zb 0.6

Zab =

Za Zb + Zb Zc + Zc Za (0.2)(0.6) + (0.2)(0.6) + (0.6)(0.6) = = 1. Zc 0.6

2.4 Circuit Theorems

59

2.4

Circuit Theorems

2.4.1

Thévenin’s Theorem and Norton’s Theorem

If the interest is to find the current and voltage in a restricted part of a circuit, then simplified procedures can be used rather than complete nodal or mesh analysis. In one of the two procedures, using Thévenin’s theorem, the rest of the circuit is replaced by an equivalent voltage source acting at the point of interest. Thévenin’s theorem is convenient in such applications as to find the load for maximum average power transfer in a circuit. In the other procedure using Norton’s theorem, the rest of the circuit is replaced by an equivalent current source acting at the point of interest. The procedures are, respectively, special cases of nodal and loop analysis.

Thévenin’s Theorem Any linear combination of voltage and current sources, independent or dependent, and resistors with two terminals can be replaced by a fixed voltage source Voc , called the Thévenin equivalent voltage, in series with a resistor Req , called Thévenin equivalent resistance. The Thévenin equivalent circuit has an equivalent volt–ampere relationship only from the point of view of the load. It is a restricted kind of source equivalence. The equivalent source is derived as follows: 1. Remove the load circuit, any combination of independent and dependent voltage and current sources, and linear and nonlinear resistors. Find the open-circuit voltage, called Voc , across the load circuit terminals. 2. Find the resistance across the load terminals, after short-circuiting all independent voltage sources and open-circuiting all independent current sources, called Req . This step can also be carried out by finding the short-circuit current Isc through the load circuit terminals and Req = Voc /Isc . Thévenin equivalent circuit is shown in Fig. 2.32a. The equivalent voltage source, replacing a circuit, is shown on the left side of the two nodes a and b. The constraint is that the current through the load circuit, represented by the resistor RL in the figure, is the same as that produced by the actual circuit. The bridge circuit, we analyzed earlier in this chapter, is shown in Fig. 2.33. The load resistor is R5 . The problem is to find the load current in the actual circuit using Thévenin’s theorem. Solution: Thévenin’s Theorem Method First, we have to find the open-circuit voltage Voc = Vab . With the load resistor removed, the current in each of the two parallel branches is 1/(3 + 1) = 0.25 A. Therefore, Va = 0.75, Fig. 2.32 Thévenin and Norton equivalent circuits

Vb = 0.25,

Req

Vab = Voc = 0.5 V.

a

a IL

Voc

RL

IL Isc

Req

b (a)

RL b

(b)

60

2 DC Circuits

Fig. 2.33 A bridge circuit with the load resistor R5

1Ω

R1

R3

R1

1Ω

R2

a 3Ω

R3

Fig. 2.35 A bridge circuit and the load resistor replaced by a current source supplying current I

b R4

3 2Ω

Req

3Ω Voc

3Ω

Va Vb I =? 1Ω 3Ω R4 1Ω

1V

Fig. 2.34 Equivalent resistance from ab and the Thévenin equivalent circuit replacing the actual circuit

R2 R5

R5

0.5V

1Ω

1Ω

IL 0.2A

V1 R1

1Ω

R2

1V

R1

3Ω

R5

1Ω V

1V

R2

3Ω

R4

1Ω

I

1Ω R3

3Ω

R4

1Ω

R3

3Ω V2

Now, we have to find the equivalent resistance, Req . For that purpose, the load resistor has to be removed and the voltage source short-circuited, as shown on the left side of Fig. 2.34. The net resistance at the terminal pair ab is the sum of the resistances of the pair of parallelly connected resistors 1 and 3 . That is, 3 1×3 1×3 + = . Req = 1+3 1+3 2 The right side of the Fig. 2.34 shows the load resistor connected to the Thévenin equivalent circuit replacing the actual circuit. The current in the load resistor is IL =

0.5 = 0.2 A, 1.5 + 1

as found in our earlier analysis by the loop method. Nodal Analysis The load current supplied by a circuit, to be replaced by an equivalent circuit, was open-circuited to find the Thévenin equivalent circuit in the first approach. Alternatively, the same effect can be achieved by inserting a bucking voltage to make the current through the load circuit zero, as shown in Fig. 2.35. Thévenin equivalent circuit is characterized by the equation

2.4 Circuit Theorems

61

V = I Req + Voc , where I is the load current. If I = 0, V = Voc , the open-circuit voltage. If V = 0, Isc = − RVoc , the eq short-circuit current. Applying KVL and KCL at the top and bottom nodes at the right side of Fig. 2.35, the equilibrium equations are V1 − V2 = 1 V1 − V V 2 V2 − V V1 + + + = 0. 3 1 1 3 The second equation is obtained using the fact that the current leaving the bottom node and entering the top node must be equal. Solving for V1 and V2 in terms of V , we get V1 =

V +1 2

and

V2 =

V −1 . 2

The equilibrium equation at node with voltage V is V − V1 V − V2 + = I. 1 3 Substituting for V1 and V2 in this equation, we get V =

3 1 I + = I Req + Voc . 2 2

The first method seems to be easier for this problem.

Norton’s Theorem Any linear combination of voltage and current sources, independent or dependent, and resistors with two terminals can be replaced by a fixed current source Isc , called Norton equivalent circuit, in parallel with a resistor Req , as shown in the right side of Fig. 2.32b. The equivalent source is derived as follows: 1. Remove the load circuit, any combination of independent and dependent voltage and current sources, and linear and nonlinear resistors, and find the short-circuit current, called Ieq or Isc , through the load circuit terminals. 2. Find the resistance across the load terminals, after short-circuiting all independent voltage sources and open-circuiting all independent current sources, called Req . This step can also be carried out by finding the short-circuit current Isc through the load circuit terminals and Req = Voc /Isc . By current division, from Fig. 2.32b, we get IL = Isc

Req Voc = . Req + RL Req + RL

62 Fig. 2.36 The short-circuit current Isc and the Norton equivalent circuit replacing the actual circuit

2 DC Circuits 2 3A 1 2A R1

1V

1 6A R2

1Ω

3Ω

Isc 0.5V 1 1 3A 2A R4 1Ω

0.5V 1 6A R3 3Ω

a 0.2A 1 3A

3 2Ω

1Ω

b Norton equivalent circuit is characterized by the equation V = I Req + Voc

or

I=

V − Isc . Req

For the bridge circuit, we have found Req = 32 , which is the same for both equivalent circuits. Finding the short-circuit by short-circuiting the load resistor is shown in the left side of Fig. 2.36. The total resistance, after short-circuiting the load resistor, becomes 3 , 2

13+13=

and

Itotal = 1/(3/2) =

2 A. 3

This current gets divided between the resistors. The current flowing through R1 = (1 − 0.5)/1 = 1/2 A. The current flowing through R3 = 0.5/3 = 1/6 A. Therefore, Isc = The load current is IL =

1 1 1 − = A. 2 6 3

    2 3 3 = = 0.2 A. 15 2 +1 2 1 3

3 2

The Norton equivalent circuit is shown on the right side of Fig. 2.36. The voltage across the load resistor in the Thévenin equivalent circuit is VL =

Voc RL . Req + RL

The voltage across the load resistor in the Norton equivalent circuit is VL =

Req Isc RL , Req + RL

which are the same as Voc = Isc Req A circuit with voltage and current sources is shown in Fig. 2.37. The problem is to find the current through the resistor R4 . In order to find Req , we have to short-circuit the voltage sources and open-

2.4 Circuit Theorems

63

Fig. 2.37 A circuit with voltage and current sources

R3

V2 I1 R1 V1

2Ω

1A V3 ×

I2

3Ω R5 4Ω

R4

1Ω

I3

×

R2 3Ω 1V

1V

Fig. 2.38 The circuit with a bucking voltage across R4

V2

R3 2Ω

R1 V1

1Ω

1A V3 = V V I R5 4 Ω

R2 3 Ω 1V 1V

circuit the current source and find the resistance across the terminals marked × in the figure, with R4 disconnected. The rightmost part of the circuit remains and the rest disconnected. Therefore, Req = 4. Let us find the open-circuit voltage across terminals, with R4 disconnected. At node 3, the equilibrium equation is V3 = −1 4

and

V3 = −4 V.

Now, Voc = −4 + 1 = −3 V. Therefore, the current through R4 is 3 Voc −3 = − A. = R4 + Req 4+3 7 The sign of Voc and Req may be positive or negative, depending on the circuit constraints. Using the alternative approach, let us insert a bucking voltage across R4 , as shown in Fig. 2.38. At the node 3, the equilibrium equation is I−

(V − 1) = 1. 4

Simplifying, we get V = 4I − 3 and

V = I Req + Voc .

64

2 DC Circuits

V1

1Ω

+ −

R1

2Ω 2(V2 −V1 )

R2 3Ω 1V

R3

R1

1Ω

R3

V2 + −

I

V2

2Ω a 2(V2 −V1 )

+ −

R3

V2

R1

1Ω

2Ω 2(V2 −V1 ) 1V

V1 1V

b

Fig. 2.39 A circuit with a controlled voltage source

A circuit with a controlled voltage source is shown in Fig. 2.39. The problem is to find the current through the resistor R2 . By nodal analysis, we get (V2 − 1) V2 V2 + 2(V2 − 1) + + = 0. 1 3 2 Solving, we get V2 = 12/17 and the current through R2 is 4/17 A. Thévenin Theorem Method Applying KVL to the middle figure, we get (1 − V2 ) + 2(1 − V2 ) = 1 + 2(V2 − 1)

and

V2 = Voc = 0.8 V.

We can find the short-circuit current Isc = 1 + 1 = 2 and Req = 0.8/2 = 0.4. There is another method to find Req . In order to find Req and the application of linearity, leave the dependent sources. Short-circuit independent voltage sources and open-circuit independent current sources. Then, apply a voltage source, such as 1 V, across the load terminals, as shown in the rightmost figure. Find the current through this source and the inverse of the current is Req . For the example, the current is 2.5 A. Therefore, Req = 1/2.5 = 0.4. Alternatively, insert a current source, such as 1 A, between the load terminals. The voltage across the source is Req . For the example, applying KCL at node 2, we get 3V2 V2 + = 1. 1 2 Solving, we get V2 = 0.4 and Req = 0.4. The current through R2 is 0.8 4 = A. 3 + (0.4) 17 By applying a bucking voltage across the load terminals and applying KCL, we get V + 2(V − 1) (V − 1) −I + =0 1 2

and

V = 0.4I + 0.8.

2.4 Circuit Theorems

65

Fig. 2.40 (a) A circuit with voltage sources; (b) equivalent circuit with current sources

0.5A

1V R2

2Ω R3

R3

V R1

R2

2Ω 3Ω

3Ω R1

1Ω

1Ω 1A

1V (a)

(b)

In circuits with both independent and dependent sources, the dependent source and its controlling variable must not be split when the circuit is broken to find the Thévenin or Norton equivalent.

Source Transformation Thévenin and Norton equivalent circuits provide the same volt–ampere relationship at the terminals. It is possible to substitute one source by another, called source transformation, to simplify the circuit analysis. Source transformation is most useful if the source is localized to some portion of the circuit. Consider the circuit shown in Fig. 2.40a with voltage sources. The problem is to find the current through R3 . Applying KCL at the middle node, we get (V − 1) V (V + 1) + + =0 1 3 2

and

V =

3 V. 11

Consider the transformed circuit shown in Fig. 2.40b with current sources. By current division, we get IR3 = 0.5

1 2 = A 11 11

and the voltage across the resistor is 3/11, as obtained above. With current sources, we got the result by using the simpler current-division formula.

2.4.2

Maximum Power Transfer Theorem

Systems absorb maximum power only for short periods. For most of the time, the load resistance is high compared with the source resistance resulting in a high efficiency. For an ideal source, Req = 0. It is often of interest in applications to find the condition for maximum power transfer from source to load. Using Thévenin’s theorem, any linear circuit can be represented by an ideal voltage source Voc in series with a resistance Req . A load resistor RL is connected across this equivalent circuit, as shown on the left side of Fig. 2.32. The problem is to find out the value of RL for a given Req . The current through RL in Fig. 2.32 is I=

Voc . RL + Req

66

2 DC Circuits

The power absorbed by RL is PL = I 2 RL =

2 Voc RL . (RL + Req )2

To find the maximum power transfer, we differentiate this expression with respect to RL and equate to zero. That is, 2 − 2V 2 R (R + R ) (RL + Req )2 Voc dPL eq oc L L = = 0, dRL (RL + Req )4 which yields RL = Req . The maximum power delivered is Pmax =

2 Voc . 4Req

Maximum power transfer theorem requires that maximum power is transferred between the source and load, when RL = Req . At maximum power transfer, half of the power is wasted in the source. However, circuits absorb maximum power only for short periods. For most of the time, the load resistance is high compared with the source resistance resulting in a low loss of power. For an ideal source, Req = 0 providing 100% efficiency of power transfer. Example Find the load resistance Rm in the bridge circuit, shown in Fig. 2.41a, for maximum power transfer from the source to the load. We have to find the Thévenin’s equivalent voltage and resistance across the load resistance terminals. Replacing the bottom half of the  circuit by Y -circuit, we get, with Zbc = 1,

Zac = 3,

Zab = 1

Za =

3×1 = 0.6 3+1+1

Zb =

1×1 = 0.2 3+1+1

Zc =

3×1 = 0.6. 3+1+1

Fig. 2.41 (a) A bridge circuit; (b) circuit reduction using  − Y transformation

R1

1Ω

R1

Rm R5

1V

a 0.6Ω

1Ω R3

3Ω

(a)

1Ω

R4

1Ω 0.6Ω c (b)

Req = Rm b 0.2Ω

2.5 Application

67

P

0.2078

0.20510 0.6

0.6364

0.8

RL Fig. 2.42 Load resistor versus power transferred

The transformed circuit, with the voltage source short-circuited, is shown in Fig. 2.41b. Now, Zeq = 0.2 + (0.6  (0.6 + 1)) = 0.2 +

(0.6 + 1)(0.6) = 0.6364. (0.6 + 1 + 0.6)

The Voc across the load terminals of the transformed circuit, with the voltage source inserted, is Voc =

(1 + 0.6)1 = 0.7273. 1 + 0.6 + 0.6

Therefore, the maximum power transferred is Pm =

2 Voc (0.72732 ) = 0.2078 W. = 4Zeq (4 × 0.6364)

Figure 2.42 shows power transferred for a range of values of the load resistance. The peak value occurs when the load resistance is equal to the Thévenin’s equivalent resistance. In power system applications, efficiency of power transmission is important, while maximum power transfer is important in signal transmission.

2.5

Application

While AC power supply is advantageous for generation, transmission, and distribution, the DC power supply is equally important in utilization of electrical energy. In some industrial applications, DC supply is preferred. In electrical traction, DC supply is extensively used. In numerous electrical and electronic appliances, the circuit works with a DC supply. Invariably, the AC input is converted to provide the required DC supply. The automobile electrical system is typical one of the many systems in day-to-day usage.

2.5.1

Strain Gauge Measurement

When we apply a force to a body, the amount of deformation of it is the strain. Strain is defined as the ratio of the increment in length to the original length. ε=

δL . L

68

2 DC Circuits

Fig. 2.43 Wheatstone bridge circuit for strain measurement

R1 V +

R2 A

R3

Rs

It can be tensile or compressive. The metallic strain gauge is a thin wire, a foil arranged in a grid pattern. It is bonded to a thin backing and attached to the test specimen. When the specimen experiences a strain, the electrical resistance of the gauge varies linearly. Typical value of the resistance is about 1000 . The sensitivity of a strain gauge, expressed as the gauge factor (GF), is the ratio of fractional change in resistance to the fractional change in length. GF =

δR/R δR/R = . δL/L ε

Typical value for GF is 2. The strain gauge is compensated suitably to reduce its effect with temperature changes. The change in resistance is typically 0.1 . In order to measure this small change accurately and also compensate for temperature changes, strain gauges are usually used in bridge circuits, called Wheatstone bridge, with voltage or current source. The Wheatstone bridge circuit for strain measurement is shown in Fig. 2.43. The variable resistance R3 consists of R and δR. Initially, δR = 0. The bridge is balanced by adjusting R so that the current through the ammeter is zero. That is R2 Rs = R. R1 When the gauge experiences strain, its value becomes Rs + δRs . Now, the bridge is unbalanced. By adjusting δR, the bridge is again balanced. Then, Rs + δRs =

R2 (R + δR) R1

and

δRs =

R2 δR. R1

With δRs , Rs and the gauge factor known, the strain is determined.

2.6

Summary

• An electric circuit, for theoretical analysis, is an interconnection of idealized representation of physical components, such as voltage and current sources, switches, resistors, inductors, and capacitors. • The relationship between the voltage across an element and the current through it is called its volt–ampere relationship and is linear in the specified operating ranges. • With the sources and circuit elements specified, the purpose of circuit analysis is to determine the voltages and currents at all parts of the circuit. • Since the current and voltage in an element are related through its volt–ampere relationship, circuits can be analyzed in terms of the branch currents alone or branch voltages alone.

2.6 Summary

69

• The method of circuit analysis based on currents is called mesh or loop analysis. The other method based on voltages is called nodal analysis. • The equilibrium conditions for a circuit can be established in either of the two ways: (1) through a set of N equations, using KVL, in which the mesh currents are the independent variables; (2) through a set of M equations, using KCL, in which the branch voltages are the independent variables. • An ideal voltage source is characterized by its volt–ampere relationship of keeping its terminal voltage same irrespective of the current drawn by the load circuit connected to it. • A circuit, geometrically, is characterized by its branches and nodes. • A loop is a closed path in a circuit. It starts at a node, passes through a set of nodes (passing through each node only once), and returns to the starting node. A loop is independent if it contains at least one branch that is not part of any other independent loop. • A tree is any set of branches of the circuit that is sufficient to connect all its nodes. A tree contains no closed paths. There is no unique tree corresponding to a circuit. • Any of the left out branches of a circuit in forming a tree are called the links. • The equilibrium state of a circuit is determined by the chosen independent link currents flowing through the selected loops. • Procedure for circuit analysis using loops involves: (1) selection of an appropriate number of independent current variables and the directions of current flow; (2) expressing the dependent current variables, by applying KCL at nodes, in terms of independent current variables; (3) applying KVL around the loops to set up a set of simultaneous equations; (4) solving for the independent currents and finding the currents in all the branches; (5) verifying the solution using KVL and KCL. • For certain type of circuits under some conditions, the general procedure can be simplified. • The selection of the direction of the currents in the branches can be arbitrary. If a current flows in the direction opposite to that assumed, the analysis result will be negative-valued. • The equilibrium state of a circuit can also be determined by a set of independent voltages, called the nodal method. • KCL is applied at nodes of the selected tree of the circuit to find the required equilibrium equations. • The determinant of the impedance matrix must be nonzero. • The determinant of the admittance matrix must be nonzero. • If a voltage source, independent or dependent, is not connected to the ground node, then its two nodes and any elements connected in parallel with it is called a supernode. • An ideal current source is characterized by its volt–ampere relationship of keeping its terminal current same irrespective of the voltage across the load circuit connected to it. • The response of a circuit to a linear combination of inputs is linear if the output is also the same linear combination of the individual outputs to the inputs. • The voltage or current provided by a controlled or dependent source, a 3-terminal device, is controlled by another voltage or current at some other part of the circuit. • Certain configurations of circuits, such as Y and  circuits, often appear in applications and it is required to transform one form from another for simpler circuit analysis. • Using Thévenin’s theorem, the current through a part of the circuit can be easily found by replacing the rest of the circuit by an equivalent voltage source acting at the point of interest. • Using Norton’s theorem, the current through a part of the circuit can be easily found by replacing the rest of the circuit by an equivalent current source acting at the point of interest. • It is possible to substitute one source by another, called source transformation, to simplify the circuit analysis. • Maximum power transfer theorem gives the condition for maximum power transfer from source to load.

70

2 DC Circuits

Exercises 2.1 Given a circuit diagram with the independent currents, find the corresponding tree and the loops. Analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. Find the power consumed by the circuit. 2.1.1 The circuit is shown in Fig. 2.44. * 2.1.2 The circuit is shown in Fig. 2.45. 2.1.3 The circuit is shown in Fig. 2.46. 2.2 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. Find the power consumed by the circuit. 2.2.1 The circuit is shown in Fig. 2.47. * 2.2.2 The circuit is shown in Fig. 2.48. 2.3 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. Find the power consumed by the circuit. 2.3.1 The circuit is shown in Fig. 2.49. * 2.3.2 The circuit is shown in Fig. 2.50. 2.4 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Find the power consumed by the circuit. Verify the results using KVL and KCL. Analyze the circuit using the linearity property also. Verify that the power consumed is equal to the power supplied. 2.4.1 The circuit is shown in Fig. 2.51. * 2.4.2 The circuit is shown in Fig. 2.52. 2.5 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. Find the power consumed by the circuit. Analyze the circuit using the linearity property also. 2.5.1 The circuit is shown in Fig. 2.53. * 2.5.2 The circuit is shown in Fig. 2.54. Fig. 2.44 Circuit for Exercise 2.1.1

I1 I2 R1

2Ω

R2

1Ω

R4

3Ω

R5

1V

1Ω R3

2Ω I3

Exercises

71

I1

Fig. 2.45 Circuit for Exercise 2.1.2

I2

R1

4Ω

1Ω

R5

1V R3

Fig. 2.46 Circuit for Exercise 2.1.3

R2

1 Ω I3 2Ω R4 3 Ω

I1 I2 R1

2Ω

R2

1Ω

R4

4Ω

R5

1V

1Ω R3

3Ω I3

Fig. 2.47 Circuit for Exercise 2.2.1

I2

I1 R1 R6

1Ω

R2

4Ω

1V R4

1Ω

R5

1Ω

2Ω R3

3Ω

I3

Fig. 2.48 Circuit for Exercise 2.2.2

I2

I1 R1 R6

1Ω

R2

3Ω

2V R4

1Ω

R5

1Ω

1Ω R3

4Ω

I3

72

2 DC Circuits

Fig. 2.49 Circuit for Exercise 2.3.1

I1

I2

R1 R3

3Ω

I3 3Ω

I1

R5

R2

1V

1Ω R3

Fig. 2.50 Circuit for Exercise 2.3.2

2Ω

R4

4Ω

I2

R1

1Ω 1V

R2

3Ω

R3

3Ω

R4

2Ω

1Ω

I3

1A

R3

Fig. 2.51 Circuit for Exercise 2.4.1

I1

3Ω

I2

I3 R4

R1 1Ω

3Ω R5 4Ω

R2 2Ω 1V 1V

Fig. 2.52 Circuit for Exercise 2.4.2

R3 I1

2Ω

R1 1Ω

1A I2

I3 R4

3Ω R5 4Ω

R2 3Ω 1V 1V

Exercises

73

R1

Fig. 2.53 Circuit for Exercise 2.5.1

I1

I2

1Ω R2

3Ω R3

2Ω 1V

3A

R1

Fig. 2.54 Circuit for Exercise 2.5.2

I1

I2

4Ω R2

3Ω R3

2Ω 1V

2A

R5

Fig. 2.55 Circuit for Exercise 2.6.1

2Ω R2

+ −

V1 I1 2(V1 − V4 )

4Ω

R1 1Ω

1V I2 R3

4Ω

V4 I3 R4

3Ω

2A

2.6 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. Find the power consumed by the circuit. 2.6.1 The circuit is shown in Fig. 2.55. * 2.6.2 The circuit is shown in Fig. 2.56. 2.7 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. Find the power consumed by the circuit.

74

R5 I1 V1

2Ω R2

+ −

Fig. 2.56 Circuit for Exercise 2.6.2

2 DC Circuits

2(V1 − V4 )

4Ω

R1 3Ω

1V

V4 I3

I2 R3

4Ω

R4

1Ω

2A

R5

Fig. 2.57 Circuit for Exercise 2.7.1

I1

2Ω R2

Ix

1Ω

R1 4Ω

R4 I2 R3

2Ω

I3 1V

3Ω

2Ix

R5

Fig. 2.58 Circuit for Exercise 2.7.2

I1

2Ω R2

Ix

1Ω

R1 1Ω

R4 I2 R3

4Ω 3Ω

I3 1V

2Ix

2.7.1 The circuit is shown in Fig. 2.57. * 2.7.2 The circuit is shown in Fig. 2.58. 2.8 Find the Thévenin and Norton equivalent circuits and determine the current through the load resistor. 2.8.1 The load resistor is R1 in Fig. 2.2a. 2.8.2 The load resistor is R3 in Fig. 2.6b. 2.8.3 The load resistor is R3 in Fig. 2.8. 2.8.4 The load resistor is R2 in Fig. 2.10a. 2.8.5 The load resistor is R3 in Fig. 2.12. 2.8.6 The load resistor is R1 in Fig. 2.15. 2.8.7 The load resistor is R3 in Fig. 2.19. 2.9 Find the value of the resistor R for maximum power transfer from the source to the resistor R. 2.9.1 The circuit is shown in Fig. 2.59. * 2.9.2 The circuit is shown in Fig. 2.60.

Exercises

75

Fig. 2.59 Circuit for Exercise 2.9.1

2Ω

3Ω

1V + 5Ω R

4Ω

R

3Ω

Fig. 2.60 Circuit for Exercise 2.9.2

1V + 1Ω 2Ω

Fig. 2.61 A circuit with current sources

4Ω

0.5A R2 2Ω R3

3Ω R1 1Ω 1A

Fig. 2.62 A circuit with voltage sources

3V R2

2Ω R3

V R1

1Ω 3Ω 2V

2.10 Find the current through the resistor R3 by analyzing the circuit as shown and also after source transformation. * 2.10.1 The circuit is shown in Fig. 2.61. 2.10.2 The circuit is shown in Fig. 2.62.

AC Circuits

3

Alternating current (AC) is also movement of electrical charge, but changes direction periodically. As the current is alternating, voltage must also alternate. That is, the polarity is changed in each cycle. The advantages of AC power supply are: 1. Power transmission and distribution is more efficient. Power has to be transmitted over long distances from source to consumption. Transmitting power at high voltages reduce transmission losses. But, power is utilized at low voltages. Transformers are used to provide the change in levels. 2. AC power generation is easier. It is also used in most high power applications. However, DC is used in majority of electrical appliances and certain industrial applications, such as electric traction. Nowadays, DC power is mostly generated by batteries, solar cells, or by converting AC to DC. Therefore, the study of both DC and AC circuits is required. The basic principles are essentially the same. However, the study of DC circuit analysis, without transient analysis, is relatively simpler, as the amplitude profile of voltages and currents is a straight line. Therefore, we study DC circuit analysis first followed by AC circuit analysis. While the DC waveform is a straight line, as can be seen on the screen of an oscilloscope, the AC waveform is sinusoidal. The well-known trigonometric sine and cosine functions are the mathematical definitions of this waveform.

3.1

Sinusoids

A sinusoidal waveform is a linear combination of sine and cosine waveforms. Sinusoidal representation of signals is indispensable in the analysis of signals and systems for the following reasons. The steady-state waveform, due to an input sinusoid, in any part of a linear system is also a sinusoid of the same frequency as that of the input differing only in its magnitude and phase. In addition, the sum of any number of sinusoids of the same frequency is also a sinusoid of the same frequency. The frequency of a sinusoid remains the same in its derivative form also. Therefore, system models, such as differential equation and convolution, reduce to algebraic equations for a sinusoidal input for linear systems. Further, due to the orthogonal property, an arbitrary signal can be decomposed into a set of sinusoids easily. In addition, this decomposition can be implemented faster, in practice, © Springer Nature Switzerland AG 2020 D. Sundararajan, Introductory Circuit Theory, https://doi.org/10.1007/978-3-030-31985-4_3

77

78

3 AC Circuits

using efficient numerical algorithms resulting in finding the system output faster than other methods. Physical systems also, such as a combination of an inductor and a capacitor, produce an output of sinusoidal nature. The motion of a simple pendulum is approximately sinusoidal. The Polar Form of Sinusoids There are two forms of representation of real sinusoids. At a given angular frequency ω, a sinusoid is characterized by its amplitude A and its phase θ (called the polar form) or by the amplitudes of its sine and cosine components (called the rectangular form). Signal amplitude can be either positive or negative. However, magnitude is always positive. The polar form of a sinusoid is x(t) = A cos(ωt + θ ),

−∞ < t < ∞.

A sinusoidal waveform has a positive peak and a negative peak in each cycle. The distance of either peak of the waveform from the horizontal axis is its amplitude A. The cosine function is periodic, as cos(ωt) = cos(ωt + 2π ). It repeats its values for t = t + T = t + 2π/ω, where T is its period in seconds. Then, the cyclic frequency of the sinusoid is f = 1/T Hz (cycles/second). The independent variable t, while time in most applications, can be anything else also, such as distance. 2π Sinusoids x(t) = cos( 2π 8 t) and x(t) = 2 sin( 8 t) are shown in Fig. 3.1a. Cosine and sine waveforms are special cases of a sinusoidal waveform. Cosine waveform has its peak value 1 at t = 0. Taking it as a reference, its phase is defined as zero radians. The radian frequency is ω = 2π/8 rad/s. Its period is T = 2π/ω = 8 s. That is, it makes one complete cycle in 8 s, as shown in the figure, and repeats indefinitely for −∞ < t < ∞. Its cyclic frequency is f = 1/8 Hz. The sine waveform x(t) = 2 sin( 2π 8 t) has its peak value 2 at t = 2. Taking the cosine waveform as the reference, its first peak occurs at t = 2, after a delay of 2 s, which is one-fourth of a cycle in the period 8. Since one complete cycle corresponds to 2π radians or 360◦ , its phase is defined as −π/2 radians or −90◦ . That is, 2 sin(

2π π 2π t) = 2 cos( t − ). 8 8 2

2

0.8660 0.5

x(t)

x(t)

1 0

0

-1 -2 0

2

4

t

6 (a)

8

-1 0

2

2π 2π Fig. 3.1 (a) x(t) = cos( 2π 8 t) and x(t) = 2 sin( 8 t); (b) xe (t) = 0.5 cos( 8 t), xo (t) = 2π π cos( 8 t − 3 )

4

6 (b)

t √

3 2

8

sin( 2π 8 t), and x(t) =

3.1 Sinusoids

79

Therefore, given a sinusoidal waveform in terms of sine waveform, it can be expressed, in terms of cosine waveform as A sin(ωt + θ ) = A cos(ωt + (θ − π2 )). Sinusoids remain the same by a shift of an integral number of their periods, as they are periodic. Similarly, A cos(ωt + θ ) = A sin(ωt + (θ + π2 )).

3.1.1

The Rectangular Form of Sinusoids

Figure 3.1b shows the sinusoid 2π π t− ) 8 3 in solid line. Its peak value of 1 occurs at t = 4/3 s. Therefore, its phase is −((4/3)/8)2π = −π/3 radians or −60◦ . Using the trigonometric subtraction formula, we get the rectangular form as x(t) = cos(

cos(

π π 2π π 2π 2π t − ) = cos( ) cos( t) + sin( ) sin( t) 8 3 3 8 3 8 √ 1 2π 2π 3 = cos( t) + sin( t). 2 8 2 8

The rectangular form expresses a sinusoid as the sum of its sine and cosine components, which are also, respectively, its odd and even components. The sine and cosine components are shown, respectively, by dotted and dashed lines in the figure. In general, we get A cos(ωt + θ ) = A cos(θ ) cos(ωt) − A sin(θ ) sin(ωt) = C cos(ωt) + D sin(ωt), where C = A cos θ

and

D = −A sin θ.

The inverse relation is A=

3.1.2

 C 2 + D2

and

C −D ). θ = cos−1 ( ) = sin−1 ( A A

The Complex Sinusoids

While the sinusoidal waveform is used in practical systems, its mathematically equivalent form, called the complex sinusoid, v(t) = V ej (ωt+θ) = V ej θ ej ωt ,

−∞ < t < ∞

is found to be indispensable for analysis due to its compact form and ease of manipulation of the exponential function. ej ωt is the complex sinusoid with unit magnitude and zero phase. The complex (amplitude) coefficient is V ej θ . The amplitude and phase of the sinusoid is represented by the single complex number V ej θ , in contrast to using two real values in the real sinusoid. Due to Euler’s identity, we get  V  j (ωt+θ) e v(t) = + e−j (ωt+θ) = V cos(ωt + θ ). 2

80

3 AC Circuits

The complex exponential functions separately have no physical significance. Their sum represents a physical voltage. However, the response of a circuit to V ej ωt yields enough information with ease to deduce the response to real sinusoids. With ω = 0 and θ = 0, v(t) = V ej (ωt+θ) = V . Therefore, AC circuit analysis is of the same form as that of the DC. The difference is that the variables are characterized by 2 values, (V , θ ) at a given frequency ω, rather than one, V , in the DC case. Note that ω = 0 and θ = 0 for DC.

3.2

AC Circuit Analysis

AC circuit analysis is carried out using sinusoidal input. The reason for that is the steady-state voltages and currents in any part of a linear circuit are of the same frequency as that of the source. The input–output relationship becomes algebraic and, hence, the circuit analysis. Any nonsinusoidal source can be decomposed into frequency components with different frequencies. The total output is the sum of the partial outputs due to sources with various frequencies. That is transform theory. Therefore, although most practical source waveforms are not sinusoidal, emphasis is given to analysis with sinusoidal inputs. Further, the real sinusoidal functions are difficult to manipulate. Therefore, circuit analysis boils down to analysis with sources of the form Aej (ωt+θ) , which is a mathematically equivalent representation of the real sinusoid. Then, the output to the real sinusoid is deduced from the analysis results.

3.2.1

Time- and Frequency-Domain Representations of Circuit Elements

In addition to the resistor used in DC circuit analysis, two more circuit elements, inductor and capacitor, are used in AC circuit analysis. Figure 3.2a, b show, respectively, the input–output relationship of an inductor of value L henries and a capacitor of value C farads in the time-domain. The voltage across a capacitor is the time integral of the current flowing through it times the reciprocal of its value. The voltage across an inductor is the time derivative of the current flowing through it times its value. When the complex exponential source is used, the circuit elements, resistors, inductors, and capacitors, are to be represented appropriately. Volt–ampere relationships of circuit elements are shown in Table 3.1. The resistance R is not a function of frequency and remains the same in both the domains. For a resistor, the impedance is Z = R, and the admittance is the reciprocal of impedance. In the case of resistor circuits, they are also called, respectively, resistance and conductance. In the Fig. 3.2 Input–output relationships of an inductor and a capacitor in the time-domain

i = L1 di v L dt

(a)

t vdt −∞

L

i = C dv dt v

1 t idt C −∞

(b)

C

3.2 AC Circuit Analysis Table 3.1 Volt–ampere relationships of circuit elements

81

Element Time-domain Frequency-domain Impedance Admittance R L C

v = Ri(t) v = L di(t) dt i=

C dv(t) dt

V = RI V = j ωLI

Z=R Z = j ωL

Y = Y =

V =

Z=

Y = j ωC

I j ωC

1 j ωC

1 R

1 j ωL

case of inductors and capacitors, the volt–ampere relationships involve differential equations in the time-domain. However, for complex exponential input only, the relationships become algebraic, as in the case of the DC circuit analysis. The exponential is the only function that is its own derivative. This is the reason for the importance of decomposing an arbitrary input into frequency components, finding the output to each component, and summing all the responses to find the total output. Specifically, in the case of inductor, with the current being ej ωt , V = j ωLej ωt . That is, V /I = j ωL = Z, the complex form of the Ohm’s law. In the case of capacitor, with the voltage being ej ωt , I = j ωCej ωt . That is, V /I = 1/(j ωC) = Z, the complex form of the Ohm’s law. The equivalent impedance of series connected impedances is the sum of the individual impedances. The equivalent admittance of parallelly connected admittances is the sum of the individual admittances. In the analysis of complicated circuits, one may need to convert repeatedly from an admittance basis to an impedance basis, and vice versa. The impedance of a series connected resistor R and inductor L is Z = R +j ωL. That is, the real part is a resistance. In general, the real part is called the resistive part and the imaginary part is called the reactive part. Both the parts may be combination of resistance, inductance, and capacitance elements. The differential equation characterizing a series RL circuit, with the input V ej ωt and the response i(t), is di(t) + Ri(t) = V ej ωt . L dt A function of the form i(t) = I ej ωt satisfies the differential equation. Therefore, the differential equation reduces to an algebraic equation, in the frequency-domain, yielding the solution I=

V V = , R + j ωL Z

where Z = R + j ωL is called the impedance of the circuit. The impedance is the opposition to the flow of current by the circuit for a complex exponential excitation. The admittance is the reciprocal of the impedance. The magnitude and the phase of the current, assuming the phase angle of V is zero, are, respectively, V ωL  ). and θ = − tan−1 ( 2 2 R R + (ωL) Differentiating a sinusoid any number of times results in the sinusoid of the same frequency with only changes in the amplitude and phase. Therefore, the differential and integral equations become algebraic equations, making it easy to find the solution. No other waveform has this property. Further, an arbitrary signal can be decomposed into sinusoidal components of various frequencies in transform analysis, as presented in later chapters. These are the major reasons for the importance of AC circuit analysis using sinusoids.

82

3 AC Circuits

3.2.2

Time-Domain Analysis of a Series RC Circuit

Let us solve a simple problem of finding the steady-state current in a series RC circuit shown in Fig. 3.3, excited by a sinusoidal voltage source, v(t) = V cos(ωt) with V = 1 V and ω = 1. The differential equation characterizing the circuit, with input V cos(ωt), is 1 C



t −∞

i(t) + Ri(t) = V cos(ωt).

Differentiating this equation and dividing both sides by R, with zero initial condition, we get 1 ωV di(t) + i(t) = − sin(ωt). dt RC R The response is expected to be of the form I cos(ωt + θ ) and its derivative with respect to t is

−I ω sin(ωt + θ ).

Substituting in the differential equation, we get (−ω sin(ωt + θ ) +

1 ωV cos(ωt + θ ))I = − sin(ωt). RC R

Expanding sine and cosine functions on the left side, we get (−ω(sin(ωt) cos(θ ) + cos(ωt) sin(θ )) + =−

1 (cos(ωt) cos(θ ) − sin(ωt) sin(θ )))I RC

ωV sin(ωt). R

Combining the terms associated with cos(ωt) and equating it to zero, we get

and

Fig. 3.3 A series RC circuit

1 (cos(ωt) cos(θ )) I = 0 RC

1/RC 1 sin(θ ) = tan(θ ) = = cos(θ ) ω ωRC

and

θ = tan−1 (

cos(t)V

(−ω cos(ωt) sin(θ )) +

1 ). ωRC

i(t) + ∼

R C

1F

2Ω

3.2 AC Circuit Analysis

83

By trigonometric definition, cos(θ ) = With x = ω and y =

1 RC ,

x , r

sin(θ ) =

y r

and

r=



x2 + y2.

we get

cos(θ ) = 

ω

and

1 2 ω2 + ( RC )

sin(θ ) = 

1 RC 1 2 ω2 + ( RC )

.

Combining the terms associated with sin(ωt) and equating, we get (ω(sin(ωt) cos(θ )) +

1 ωV (sin(ωt) sin(θ )))I = sin(ωt). RC R

Simplifying, we get (ω cos(θ ) +

ωV 1 sin(θ ))I = . RC R

Substituting for cos(θ ) and sin(θ ), we get ωV V R I= = . 1 2 1 2 ω2 + ( RC ) ) R 2 + ( ωC

With V = 1, R = 2, C = 1, ω = 1, I = 0.4472, θ =  (0.4636) rad For RL circuit, with V = 1, R = 2, L = 1, ω = 1, I = 0.4472, θ =  (−0.4636) rad The input and the response are shown in Figs. 3.4 and 3.5. The current leads the voltage in the RC circuit. The current lags the voltage in the RL circuit. 1

x(t)

0.4472 0

0

5.8196

t Fig. 3.4 Excitation and response of a series RC circuit

84

3 AC Circuits

1

x(t)

0.4472 0

0

0.4636

6

t Fig. 3.5 Excitation and response of a series RL circuit

3.2.3

Frequency-Domain Analysis of a RC Circuit

When the independent variable of signals, such as time t, is not frequency, then that representation is called the time-domain representation. On the other hand, when the independent variable of signals is frequency, such as f and ω, then that representation is called frequency-domain representation. In the frequency-domain, circuit analysis becomes algebraic, rather than solving a differential equation in the time-domain. Transforms, such as Fourier transform and Laplace transform, are used to transform signals and systems from one domain into another. The differential equation characterizing the RC circuit, with input V ej ωt , is 1 C



t −∞

i(t) + Ri(t) = V ej ωt .

Differentiating both sides and assuming that the initial condition is zero, we get 1 j V ω j ωt di(t) + i(t) = e . dt RC R The response is expected to be of the form I ej ωt and its derivative is j ωI ej ωt . Substituting for i(t) in the differential equation, we get ( and I=

j V ω j ωt 1 + j ω)I ej ωt = e RC R

RC j ωCV V V jV ω V = = = = . j 1 R 1 + j ωRC 1 + j ωRC Z R + j ωC R − ωC

The magnitude and phase of I are the same as those obtained by the time-domain method. Comparing the time-domain and frequency-domain analyses, the simplicity of the latter is obvious. The steps involved in transform analysis of AC circuits are: 1. Transform the circuit from the time-domain to the frequency-domain. 2. Find the required output using nodal or mesh analysis. 3. Transform the solution back to the time-domain.

3.2 AC Circuit Analysis

85

The first task in formulating the equilibrium equations for an electric circuit is to select a set of variables, voltages or currents. They must be independent and adequate to describe the state of the network at any instant. They must be uniquely and reversibly related to all the branch variables. After selecting an appropriate set of variables, we have to use Kirchhoff’s laws. The volt–ampere relations of the circuit elements are required at this stage. Of course, voltage or current sources are required to energize a circuit and their characteristics must be taken into account. Further, we used circuit theorems in analyzing circuits. All these aspects presented for DC circuits remain the same for AC circuit analysis. One important difference in AC circuit analysis is that the independent variable is frequency. At any given frequency, a sinusoid is characterized by two parameters, amplitude and phase. These two parameters are combined into a complex quantity, called a phasor, for the convenience of analysis. A phasor is a two-element vector that represents the amplitude and phase of sinusoidal voltages and currents. Therefore, complex quantities are involved in the analysis of AC circuits. Otherwise, the analysis is similar to that of DC circuits. Same equilibrium equations are applicable for DC and AC circuits of the same structure with the circuit elements denoted by Z1 , Z2 , etc. The circuit elements for DC circuits are resistors, while those for AC circuits are resistors, inductors and capacitors. This difference comes into play only when the values of the elements are substituted in the same set of equations for the same circuit structure. We use real arithmetic in the analysis of DC circuits, while complex arithmetic is required for AC circuit analysis. Therefore, the essential part of the procedure remains the same for DC and AC circuits. Therefore, AC circuit analysis is no more complex than that of DC circuit analysis, except that of using complex arithmetic. Of course, visualization of DC voltages and currents is easier. A suggested procedure for feeling comfortable and getting used to AC circuit analysis is as follows: Denote each circuit element by Z1 , Z2 , etc. With the same structure of the circuit, use resistors with suitable values, such as 1 and 2  for Z1 , Z2 , etc. Replace the AC sources by DC sources. Write down the equilibrium equations for the type of analysis desired (nodal or mesh). Solve the equations. Verify the solution by applying KVL and KCL. Then, use the same equilibrium equations with the appropriate impedance values and AC sources. Finally, verify the solution by applying KVL and KCL. The process is similar to real and complex number arithmetic. We are more used to real arithmetic but the basic principle behind both arithmetic are the same. Another inherent property of AC circuits is that the circuit must be analyzed at each frequency individually, if sources with more than one frequency are present in the circuit. The value of the impedance changes with the frequency of the source. The response of the circuit at all frequencies must be added to find the total response. Therefore, we analyze circuits with the same structure, studied in Chap. 2, with resistors, inductances, and capacitors, and AC sources, so that the differences in DC and AC circuit analyses are clearly highlighted. As the sinusoidal waveform is everlasting, using a sinusoid as a source will result in the steadystate analysis. In practice, we switch on the sinusoidal source at some finite time, not from at some remote time. Due to this, there will be a transient response. For stable systems, such as the circuits, we study that the transient response will die down in a short time after switching a sinusoidal source, leaving only steady-state response. Transient response will be presented later. For proper steady-state analysis, ensure that the effective transient response time of the circuit is short enough compared with that of the period of the excitation.

3.2.4

Impedances Connected in Series

AC circuit analysis is also based on the same concepts for the DC analysis. The difference is that the source, the sinusoid, is characterized, at a particular frequency, by two parameters, magnitude and phase, rather than the magnitude alone in DC analysis. Further, the impedances of inductors

86

3 AC Circuits

L

+ +

VZL

-

2H +

+ ∼

2cos(2t)V

V + ∼

VZL 2cos(2t)V

+

VZC VZC 1F C ZL = jωL ZL = j4 Z = 1 Z = −j0.5 3Ω C ZR = R C jωC + + VZR VZR ZR 3 VZR = V ZL +ZC +ZR VZR = 2 3+j3.5 = 0.8471−j0.9882 j4 ZL VZL = V ZL +ZC +ZR VZL = 2 3+j3.5 = 1.3176+j1.1294 VZC = V

ZC ZL +ZC +ZR

VZL 2H

VZC 1F ZL = j4 Z = −j0.5 3Ω C VZR 3 −2 3+j3.5 = −0.8471+j0.9882 j4 −2 3+j3.5 = −1.3176−j1.1294 ∼

−j0.5 −j0.5 VZC = 2 3+j3.5 = −0.1647−j0.1412 −2 3+j3.5 = 0.1647+j0.1412

(a)

(b)

(c)

Fig. 3.6 Series circuit with a voltage source

and capacitors are also characterized, at a particular frequency, by two parameters, magnitude and phase, rather than the magnitude alone for the resistor. Due to these reasons, AC circuit analysis, while based on the same concepts, requires more arithmetic. AC circuit analysis is also indispensable since AC is widely used. In addition, most of the signals occurring in practice have arbitrary amplitude profile and it is difficult to analyze circuits with them. It becomes a necessity to decompose the signals in terms of well-defined basis signals and the sinusoid is the most suitable for that purpose. In practice, the desired circuit elements are often not available and we have to use a combination of more than one element as an equivalent one. Impedances can be combined in series and/or parallel configurations, exactly as in resistor circuits. Resistors, inductors, and capacitors have two terminals and, therefore, they come under the class of two-terminal devices or elements. In a series connection, one, and only one, terminal of an element is connected to adjoining elements. Figure 3.6a shows a resistor, an inductance and a capacitor connected in cascade, called a series circuit. A circuit is an interconnection of elements. The determination of currents and voltages at all parts of the circuit is the essence of circuit analysis. When impedances are connected in series, the voltage across them increases, with the same current flowing through them. It is similar to connecting hoses to make a longer hose. The combined impedance is the sum of all the impedances. That is, with N number of impedances connected in series, the equivalent impedance Zeq of the series circuit is Zeq = Z1 + Z2 + · · · + ZN . The same current I pass through all the impedances. Therefore, the voltage V across the series connection is V = I Z1 + I Z2 + · · · + I ZN = I Zeq . The equivalent impedance remains unchanged, irrespective of the order in which they are connected. Obviously, if all of them have the same value, then Zeq = N Z. The source voltage applied across them gets divided in proportion to their individual values. The current through the series circuit is I=

V Zeq

3.2 AC Circuit Analysis

87

and the voltage across any impedance Zn is Vn =

V Zn . Zeq

With just two impedances, Z1 and Z2 , and V , the applied voltage, in the series connection, we get V1 =

V Z1 Z1 + Z2

and

V2 =

V Z2 . Z1 + Z2

Consider the circuit in Fig. 3.6a. The circuit is energized by a voltage source V . The impedances corresponding to resistor R, inductor L, and capacitor C, respectively, are ZR = R,

ZL = j ωL,

ZC =

1 . j ωC

The impedances of the inductor and capacitor are a function of the frequency. The equivalent impedance is 1 Zeq = R + j ωL + . j ωC The voltage across the impedances are, respectively, VZR = V

ZR , ZL + ZC + ZR

VZL = V

ZL , ZL + ZC + ZR

VZC = V

ZC . ZL + ZC + ZR

Alternatively, we can find the current through the series circuit I=

V Zeq

and VZR = I ZR ,

VZL = I ZL ,

VZC = I ZC .

According to KVL, V = VZR + VZL + VZC . Consider the circuit in Fig. 3.6b. The circuit is energized by a voltage source of V = 2 cos(2t)V . The magnitude of the source is 2 V and its frequency ω = 2 rad/s. An ideal voltage source maintains a constant voltage at its terminals, irrespective of the current drawn from it. A voltage source is a constraint, clamping the voltage at certain point in the circuit. The equivalent impedance is Zeq = 3 + j 2 × 2 +

1 = 3 + j 4 − j 0.5 = 3 + j 3.5. j2 × 1

The voltage source is the real part of the complex exponential 2ej 2t = 2 cos(2t) + j 2 sin(2t). Therefore, with the understanding that only the real parts of resulting complex expressions in the circuit analysis are to be considered, we use the exponential 2ej 2t as the source voltage instead of 2 cos(2t) for mathematical convenience. As the response is of the same form throughout the circuit, the

88

3 AC Circuits

2 1.7354

v(t)

x(t)

1.3016

vr(t) i(t)

0.4339 0.2169 0

vc(t)

vl(t)

0

0.4311

1.2165

3.1416

t Fig. 3.7 Source and responses of a series RLC circuit

complex exponentials cancel out and we use its coefficient 2, called the phasor, in the manipulations. Phasor is a rotating vector representing an AC sinusoidal voltage or current. As it is a complex number, it may be written in exponential, polar, or rectangular form. The voltage drops across the impedances are ZR 2×3 = 0.8471 − j 0.9882, = VZR = V Zeq 3 + j 3.5 VZL = V

ZL 2 × j4 = 1.3176 + j 1.1294, = Zeq 3 + j 3.5

VZC = V

ZC 2 × (−j 0.5) = −0.1647 − j 0.1412 = Zeq 3 + j 3.5

The three voltage drops add up to 2 V, the source voltage, satisfying KVL around the only loop of the circuit. Alternatively, V 2 = 0.2824 − j 0.3294 I= = Zeq 3 + j 3.5 and VZR = I ZR = (0.2824 − j 0.3294)3 = 0.8471 − j 0.9882, VZL = I ZL = (0.2824 − j 0.3294)(j 4) = 1.3176 + j 1.1294, VZC = I ZC = (0.2824 − j 0.3294)(−j 0.5) = −0.1647 − j 0.1412 Figure 3.7 shows the real sinusoidal waveforms of the voltages and the current in the circuit. Although we use the complex frequency-domain representation of the circuit in its analysis, for mathematical convenience, it should be always remembered that the voltages and currents in practical circuits are real sinusoids. The real sinusoidal form of the voltages and the current are obtained by taking the real part of their corresponding form obtained in the frequency-domain analysis. i(t) = |I | cos(2t +  I ) = 0.4339 cos(2t − 49.3987◦ ) vR (t) = |VZR | cos(2t +  VZR ) = 1.3016 cos(2t − 49.3987◦ ) vL (t) = |VZL | cos(2t +  VZL ) = 1.7354 cos(2t + 40.6013◦ ) vC (t) = |VZC | cos(2t +  VZC ) = 0.2169 cos(2t − 139.3987◦ )

3.2 AC Circuit Analysis

89

1

0

1.3176+j1.1294

Imaginary

Fig. 3.8 The voltages and the currents represented in the complex plane

Real -0.1647-j0.1412 0.2824-j0.3294

0.8471-j0.9882

-1 0

1

The current through and voltage across the resistor have the same phase. That is, their positive peaks occur at the same instant. The current through the inductor lags the voltage across it. Choosing the current waveform as the reference, the nearest positive peak of the voltage drop occurs earlier in a period. The current through the capacitor leads voltage across it. The positive peak of the voltage drop occurs later in a period. The voltages and the current are represented in the complex plane in Fig. 3.8. While absolute and angle values can be computed using formulas or calling functions, it is recommended that they are visualized in the complex plane. It helps understanding and also serves as a check on numerical computation. For example, the complex value corresponding to the current is located in the middle of the bottom right quadrant. That is, the angle is about −45◦ and the exact value is −49.3987◦ . The phase of the voltage across the resistor is also the same. The absolute value can be estimated using a scale. Similarly, the other values can be estimated. Further, the results of complex arithmetic operations can also be estimated graphically, as shown in Appendix B. While we checked KVL using the complex values, it can be checked using the real sinusoids also. The sum of the voltages, in rectangular form of sinusoid, are (0.8471 + 1.3176 − 0.1647) cos(2t) = 2 cos(2t) and

(0.9882 − 1.1294 + 0.1412) sin(2t) = 0

That is, the sine component is zero, while the magnitude of the cosine component is 2. Consider the circuit shown in Fig. 3.6c. The circuit is energized by the same voltage source in Fig. 3.6b, but with the polarities reversed. The voltages and the current are the same with their phases added or subtracted by π radians or 180◦ . Consider the circuit shown in Fig. 3.9a. The circuit is energized by a current source of I A. An ideal current source maintains a constant current at its terminals, irrespective of the voltage across its terminals. A current source is a constraint, clamping the current at certain point in the circuit. As the same current flows through the impedances, the respective voltage drops are VZR = I ZR ,

and

VZL = I ZL ,

and

VZC = I ZC .

An important step in the frequency-domain analysis of circuits is to convert the real sinusoidal sources into the mathematically equivalent complex sinusoidal sources. The conversion formulas are V cos(ωt + θ ) → V  θ

90

3 AC Circuits

L

+

VZL

-

2H +

C ZL = jωL Z = 1 ZR = R C jωC + VZR VZR = IZR VZL = IZL

VZC

+

+ ∼

VZC 1F ZL = j2 Z = −j1 3Ω C + VZR VZR = IZR = −j3 VZL = IZL = 2

−sin(t)A

I + ∼

VZL sin(t)A

+

VZC = IZC = −1

VZC = IZC (a)

VZL 2H -

∼

1F ZL = j2 Z = −j1 3Ω C VZR VZR = IZR = j3 VZL = IZL = −2

VZC

VZC = IZC = 1

(b)

(c)

Fig. 3.9 Series circuit with a current source

and

π ). 2 These formulas hold both for current and voltage sources. For example, V sin(ωt + θ ) → V  (θ −

i(t) = 2 cos(ωt − and v(t) = 3 sin(ωt +

π π ) → 2 − 3 3

π π π ) → 3 ( − ). 6 6 2

The last equation is due to the fact that sin(θ ) = cos(θ −

π ). 2

Consider the circuit in Fig. 3.9b. The circuit is energized by a current source of i(t) = sin(t)A. The magnitude of the source is 1 A and its frequency ω = 1 rad/s. In the frequency-domain, the corresponding current source is I = 1 −

π π π = cos( ) − j sin( ) = −j. 2 2 2

The respective voltage drops are VZR = (−j ) × 3 = −j 3V

and

VZL = (−j ) × j 2 = 2V

VZC = (−j ) × (−j 1) = −1V . In the time-domain, the values are, respectively, vR (t) = 3 sin(t),

vL (t) = 2 cos(t),

vC (t) = − cos(t).

The equivalent impedance of the circuit is 3 + j 2 − j 1 = 3 + j 1. The total voltage drop across the circuit is (−j )(3 + j 1) = 1 − j 3. The sum of the individual drops −j 3 + 2 − 1 = 1 − j 3 is equal to this value.

3.2 AC Circuit Analysis

91

Consider the circuit in Fig. 3.9c. The circuit is energized by a current source of i(t) = − sin(t)A. As the direction of current flow is reversed, the polarities of the voltage drops are also reversed.

3.2.5

Impedances Connected in Parallel

In parallel connection, one terminal of all the elements is connected to one node and the other terminals are connected to another node. Therefore, the voltage across each element is equal. Figure 3.10 shows a resistor, an inductor, and a capacitor connected in parallel, called a parallel circuit. When elements are connected in parallel, while the voltage across all of them is the same, different currents flow through them, unless some or all of them are the same. It is similar to connecting hoses to make a wider hose. The length remains the same, but the flowing capacity increases. The combined admittance, the reciprocal of the impedance Y = 1/Z, is the sum of all the admittances. That is, with N number of elements connected in parallel, the equivalent admittance Yeq of the parallel circuit is Yeq = Y1 + Y2 + · · · + YN

and

Zeq =

1 , Yeq

where Yn = 1/Zn . The value of Zeq will be smaller than the smallest of the impedances in the parallel connection, since the total current is more. The same voltage V is applied across all the elements. Therefore, the total current I flowing through the parallel connection is I = V Y1 + V Y2 + · · · + V YN = V Yeq . The equivalent admittance remains unchanged, irrespective of the order in which they are connected. Obviously, if all of them have the same value, then Yeq = N Y . The total current gets divided in proportion to their individual admittance values. The total current through the circuit is I = V Yeq and the current through any admittance Yn is

IA ∼

IR

IL

R

L

YR =

IC C

1 R

IR = I YR +YYR L +YC

YC = jωC 1 YL = jωL

I Yn . Yeq

sin(t)A

In =

IL

IR ∼

3Ω

IC

2H 1F

0.3333 IR = (−j) 0.3333−j0.5+j = −0.4615−j0.3077

IL = I YR +YYLL +YC

−j0.5 IL = (−j) 0.3333+j0.5 = −0.4615 + j0.6923

IC = I YR +YYCL +YC

j = 0.9231−j1.3846 IC = (−j) 0.3333+j0.5

(a) Fig. 3.10 Parallel circuit with a current source

(b)

92

3 AC Circuits

IL

R

L

YR =

IC C YC = jωC 1 YL = jωL

1 R

IR = V Y R

IL

IR sin(t)V

V ∼

IR

∼

3Ω

IC

2H 1F

IR = (−j)0.3333 = −j0.3333

IL = V Y L

IL = (−j)(−j0.5) = −0.5

IC = V Y C

IC = (−j)j = 1

(a)

(b)

Fig. 3.11 Parallel circuit with a voltage source

With just two admittances, Y1 and Y2 , and I , the total current, we get I1 =

I Y1 Y1 + Y2

and

I2 =

I Y2 . Y1 + Y2

Consider the circuit in Fig. 3.10a. The circuit is energized by a current source of V = I A. The admittances are 1 1 . YR = , YC = j ωC YL = R j ωL The currents are IR = I

YR YR + YL + YC

IL = I

YL YR + YL + YC

IC = I

YC . YR + YL + YC

Consider the circuit in Fig. 3.10b. The circuit is energized by a current source of I = sin(t)A. The magnitude of the source is 1 A and its frequency ω = 1 rad/s. In the frequency-domain, the source is represented by −j . The currents through the three elements are shown in the figure. The total current adds up to −j 1, which is equal to the source current. The voltages across all the elements must be the same. That is, IL IC IR = = = −1.3846 − j 0.9231. YR YL YC Consider the circuit in Fig. 3.11a. The circuit is energized by a voltage source of V . The currents through the impedances are IR = V YR ,

IL = V YL ,

IC = V YC .

Consider the circuit in Fig. 3.11b. The circuit is energized by a voltage source of sin(t)V . The currents through the impedances are IR = (−j )0.3333 = −j 0.3333,

IL = (−j )(−j 0.5) = −0.5,

IC = (−j )j = 1.

The total admittance is 0.3333 − j 0.5 + j = 0.3333 + j 0.5. The current through the circuit is (−j 1)(0.3333 + j 0.5) = 0.5 − j 0.3333, which is equal to the sum of the individual currents through the elements.

3.2 AC Circuit Analysis

3.2.6

93

Impedances Connected in Series and Parallel

The analysis of series and parallel circuits is relatively straightforward. In general, most circuits are a combination of series and parallel circuits or connected in a random configuration in which none of the elements is in series or parallel. Obviously, combinations of the concepts of series and parallel circuits are used to analyze series-parallel circuits. Analysis of circuits with random configurations are presented in the next section. First, we have to identify the parts of the circuit with series and parallel configurations and simplify them separately. Now, the circuit gets reduced to a simpler form. These steps must be repeated until we can determine the source current. Then, using the voltage-division and current-division laws governing the series and parallel circuits repeatedly, we find the voltages and currents at all parts of the circuit. Consider the circuit in Fig. 3.12a. The circuit is energized by a voltage source, V = cos(2t + π6 )V . The magnitude of the source is 1V and its frequency ω = 2 rad/s. In the frequency-domain, it is 1 π6 = 0.866 + j 0.5. First, we have to find the combined impedance of the circuit, which is Zeq = ZC + (ZR  ZL ) = ZC +

1 = 0.6897 − j 3.2759. YR + YL

Now, the current drawn from the source is I = IC =

0.866 + j 0.5 = −0.0929 + j 0.2839A. 0.6897 − j 3.2759

This current gets divided between YR and YL . Using current-division formula, we get IR =

I YR = −0.1107 + j 0.0071 A YR + YL

IL =

I YL = I − IR = 0.0179 + j 0.2768 A. YR + YL

Applying KCL at the common point of the three elements, we get −IC + IL + IR = 0. The current through the circuit multiplied by the equivalent impedance is (−0.0929 + j 0.2839)(0.6897 − j 3.2759) = 0.866 + j 0.5,

IL

IR + ∼

5Ω

1H

YR = 0.2 IC 0.1F

(a)

1 YL = jωL = −j0.5

1 ZC = jωC = −j5

sin(2t+ π6 )A

cos(2t+ π6 )V

which is equal to the source voltage. IL

IR ∼

5Ω

1H

YR = 0.2 IC 0.1F

1 YL = jωL = −j0.5

1 ZC = jωC = −j5

(b)

Fig. 3.12 Impedances connected in series and parallel. (a) with a voltage source; (b) with a current source

94

3 AC Circuits

Consider the circuit in Fig. 3.12b. I = sin(2t + π6 )A. The magnitude of the current source is 1 and its frequency ω = 2 rad/s. In the frequency-domain, it is 1 − π3 = 0.5 − j 0.866. The source current flows through C and, therefore, I = IC = (0.5 − j 0.866)A. This current gets divided between YR and YL . Using current-division formula, we get IR =

I YR = (0.3676 + j 0.0530)A YR + YL

IL =

I YL = I − IR = (0.1324 − j 0.9190)A. YR + YL

Applying KCL at the common point of the three elements, we get −IC + IL + IR = 0. The voltage across R and L must be the same. (0.3676 + j 0.0530)/0.2 = (0.1324 − j 0.9190)/(−j 0.5) = 1.8380 + j 0.2648

3.2.7

Analysis of Typical Circuits

Circuit 3.1 Consider the bridge circuit shown in Fig. 3.13a. The structure of the circuit is the same as that we analyzed in Chap. 2. The first difference is that the DC voltage source has been changed to AC source v(t) = cos(t + π/6). The five branches of the bridge are combinations of resistors, inductance, and capacitances rather than resistors only in the circuit for DC analysis. In AC circuit analysis, the circuit has to be transformed to the frequency-domain as shown in Fig. 3.13b so that we follow essentially the same procedure for DC analysis, although elements characterized by differential equations are part of the circuit. The Source The source v(t) = cos(t + π/6) is replaced by π

π

v(t) = ej (t+ 6 ) = ej 6 ej t . I1 V1

cos(t+ π6 )V

0.1F + ∼

C1

3Ω R2

−j10 L2 1H

R5 R4 1Ω

0.1F

π 6V

∼

Z2

V2 Z3 j1

C4

(a) Fig. 3.13 (a) A bridge circuit; (b) its frequency-domain version

j3 3+j

Z1 1

1Ω L3 1H

I2

(b)

Z5 1 −j10 1−j10 I3

V3 Z4

3.2 AC Circuit Analysis

95

The magnitude and phase of the complex exponential is only required in the analysis, as ej t remains unchanged throughout the circuit. The frequency ω is 1. The use of complex exponential source makes the analysis compact and easier. As the complex exponential has a real part and an imaginary part π

v(t) = ej (t+ 6 ) = cos(t +

π π ) + j sin(t + ) 6 6

and our input is the cosine waveform, the real part is implied. Therefore, the complex coefficient of the exponential V = 1 π6 represents the input source voltage in Fig. 3.13b.

The Impedances Using the formulas in Table 3.1 for volt–ampere relationships of elements and formulas for series and parallelly connected impedances, we get, with ω = 1 of the source, Z1 =

1 R2 j ωL2 j3 = −j 10, Z2 = = , Z3 = j ωL3 = j 1, j ωC1 R2 + j ωL2 3 + j1 Z4 =

1 R4 j ωC 4

R4 +

1 j ωC4

=

−j 10 , Z5 = R5 = 1, 1 − j 10

shown in Fig. 3.13b. When computing the impedance values, it is important to take into account the frequency of the voltage or current source. For each source frequency, the analysis has to be repeated. Mesh Analysis The impedance concept enables us to use the same methods, those are applicable to DC circuits, to analyze AC circuits in the steady state. The Ohm’s and Kirchoff’s laws, in terms of complex voltage, current, and impedance, are equally applicable to AC steady-state analysis. Therefore, for an AC circuit with a similar geometrical structure as that of a DC circuit, the equilibrium equations of the DC circuit, for both mesh and nodal methods of analysis, remain the same. The difference is that the elements are characterized by real quantities for DC circuits, while they are characterized by complex quantities for AC circuits. Therefore, the formal procedure of DC circuit analysis remains the same for the AC case in steady state. The resistance values are replaced by impedance values and the DC sources are replaced by AC sources. The equilibrium equations are of the same form as those we derived for the DC bridge with the source changed and are π (3.1) Z1 (I1 − I2 ) + Z3 (I1 − I3 ) = 1 6 −Z1 (I1 − I2 ) − Z5 (I3 − I2 ) + Z2 I2 = 0 (3.2) −Z3 (I1 − I3 ) + Z5 (I3 − I2 ) + Z4 I3 = 0. Simplifying, we get

(3.3)

√ (Z1 + Z3 )I1 − Z1 I2 − Z3 I3 = (

1 3 +j ) 2 2

(3.4)

−Z1 I1 + (Z1 + Z2 + Z5 )I2 − Z5 I3 = 0

(3.5)

−Z3 I1 − Z5 I2 + (Z3 + Z4 + Z5 )I3 = 0.

(3.6)

96

3 AC Circuits

The equilibrium equations, in matrix form, are ⎡ ⎢ ⎣

(Z1 + Z3 ) −Z1 −Z3

⎤ ⎤ ⎡ √3 ( 2 + j 12 ) I1 ⎥ ⎢ ⎥ (Z1 + Z2 + Z5 ) −Z5 ⎦ ⎣ I2 ⎦ = ⎣ 0⎦ I3 −Z5 (Z3 + Z4 + Z5 ) 0 −Z1

With Z1 = −j 10,

Z2 =

j3 , 3 + j1

−Z3

Z3 = j 1,

⎤⎡

Z4 =

−j 10 , 1 − j 10

Z5 = 1,

the determinant of the impedance matrix is 27.8624 − j 7.0762. Solving the equilibrium equations, we get ⎤ ⎤ ⎡ 0.0000 − j 9.000 0.0000 + j 10.000 0.0000 − j 1.000 ⎤−1 ⎡ √3 ( 2 + j 12 ) I1 ⎥ ⎢ ⎥ ⎣ I2 ⎦ = ⎢ ⎣ 0.0000 + j 10.000 1.3000 − j 9.100 −1.0000 + j 0.000 ⎦ ⎣ 0⎦ I3 0.0000 − j 1.000 −1.0000 + j 0.000 1.9901 + j 0.901 0 ⎡

⎡

0.4750 − j 0.4873 0.4656 − j 0.5601 0.3813 − j 0.2154

⎤⎡

⎥⎢ ⎢ = ⎣ 0.4656 − j 0.5601 0.4605 − j 0.5259 0.4142 − j 0.2178 ⎦ ⎣ 0.3813 − j 0.2154 0.4142 − j 0.2178 0.7104 − j 0.2395 ⎤ ⎡ 0.6550 − j 0.1845 ⎥ ⎢ = ⎣ 0.6833 − j 0.2523 ⎦ 0.4379 + j 0.0041 {IZ1 = −0.0283 + j 0.0677,

IZ3 = 0.2171 − j 0.1886,

(

√ 3 2

+ j 12 )

⎤

⎥ 0⎦ 0

IZ5 = −0.2454 + j 0.2564}

Verifying the Solutions Applying KVL around the loops, we get IZ1 Z1 + IZ3 Z3 = (−0.0283 + j 0.0677)(−j 10) + (0.2171 − j 0.1886)(j 1) √ 1 3 +j ) = (0.6774 + j 0.2829) + (0.1886 + j 0.2171) = ( 2 2 −I2 Z2 + IZ5 Z5 + IZ1 Z1 = −(0.6833 − j 0.2523)(

j3 ) + (−0.2454 + j 0.2564)(1) + (−0.0283 + j 0.0677)(−j 10) 3+j

= (−0.4320 − j 0.5393) + (−0.2454 + j 0.2564) + (0.6774 + j 0.2829) = 0 IZ5 Z5 + I3 Z4 − IZ3 Z3 = (−0.2454 + j 0.2564)(1) + (0.4379 + j 0.0041)(

−j 10 ) − (0.2171 − j 0.1886)(j 1) 1 − j 10

= (−0.2454 + j 0.2564) + (0.4340 − j 0.0393) − (0.1886 + j 0.2171) = 0

3.2 AC Circuit Analysis

97

While these verifications in complex form are correct, as the given excitation is a real sinusoid, we have to relate the excitation to the branch voltages and currents. Consider the loop involving Z1 , Z3 and the excitation. √ 3 1 IZ1 Z1 + IZ3 Z3 = (0.6774 + j 0.2829) + (0.1886 + j 0.2171) = ( + j ). 2 2 In terms of complex exponentials, this equation corresponds to π

0.7341ej (t+0.3956) + 0.2876ej (t+0.8554) = ej (t+ 6 ) . The magnitudes and phases are the absolute values and angles of the corresponding complex coefficients. By taking the real parts, this equation corresponds to, in terms of real sinusoids, 0.7341 cos(t + 0.3956) + 0.2876 cos(t + 0.8554) = cos(t +

π ). 6

Expressing the sinusoids in rectangular form, we get (0.6774 cos(t) − 0.2829 sin(t)) + (0.1886 cos(t) − 0.2171 sin(t)) = (0.866 cos(t) − 0.5 sin(t)) In a similar manner, we can interpret the other complex quantities in terms of real sinusoids. Verifying the solution using KCL at nodes, we get I1 − IZ1 − I2 = (0.6550 − j 0.1845) − (−0.0283 + j 0.0677) − (0.6833 − j 0.2523) = 0 IZ1 − IZ3 − IZ5 = (−0.0283 + j 0.0677) − (0.2171 − j 0.1886) − (−0.2454 + j 0.2564) = 0 IZ5 + I2 − I3 = (−0.2454 + j 0.2564) + (0.6833 − j 0.2523) − (0.4379 + j 0.0041) = 0 IZ3 − I1 + I3 = (0.2171 − j 0.1886) − (0.6550 − j 0.1845) + (0.4379 + j 0.0041) = 0 Nodal Analysis Except for the differences pointed out for loop analysis, the equilibrium equations are of the same form as those derived for the DC bridge in Chap. 2. V1 =

1

√ 1 π 3 =( +j ) 6 2 2

(V2 − V1 ) (V2 − V3 ) V2 + + =0 Z1 Z5 Z3

(3.7)

V3 − V1 (V3 − V2 ) V3 + + = 0. Z2 Z5 Z4

(3.8)

The first equation is the application of KCL at the left middle node. The second equation is the application of KCL at the right middle node. Simplifying, we get (Z3 Z5 + Z1 Z3 + Z1 Z5 )V2 − Z1 Z3 V3 = Z3 Z5 V1

(3.9)

−Z2 Z4 V2 + (Z4 Z5 + Z2 Z4 + Z2 Z5 )V3 = Z4 Z5 V1 .

(3.10)

98

3 AC Circuits

With Z1 = −j 10, 

Z2 =

j3 , 3+j

Z3 = j 1,

10.0000 − j 9.0000 −10.0000 + j 0.0000 −0.3861 − j 0.8614



1.6762 + j 1.6624

Z4 = V2 V3

 =

−j 10 , Z5 = 1 1 − j 10   −0.5000 + j 0.8660 0.9070 + j 0.4093

.

The determinant of the admittance matrix is 27.8624 − j 7.0762. Solving the equilibrium equations, we get

V2 V3



 =

 =

−1 

−0.5000 + j 0.8660



−0.3861 − j 0.8614

1.6762 + j 1.6624 0.9070 + j 0.4093   −0.5000 + j 0.8660 0.0423 + j 0.0704 0.3372 + j 0.0856

0.0056 + j 0.0323 0.4142 − j 0.2178 

=

10.0000 − j 9.0000 −10.0000 + j 0.0000

0.1886 + j 0.2171

0.9070 + j 0.4093



0.4340 − j 0.0393

{VZ1 = 0.6774 + j 0.2829,

VZ2 = 0.4320 + j 0.5393,

VZ5 = −0.2454 + j 0.2564}.

Circuit 3.2 Consider the loop analysis of the circuit shown in Fig. 3.14a. The corresponding frequency-domain circuit is shown in Fig. 3.14b. The equilibrium equations are of the same form as those derived for the corresponding to DC bridge in Chap. 2 with the source changed. Z1 (I1 − I2 ) + Z3 (I1 − I3 ) + Z6 I1 = 0

(3.11)

π −Z1 (I1 − I2 ) − Z5 (I3 − I2 ) + Z2 I2 = 1 3

(3.12)

−Z3 (I1 − I3 ) + Z5 (I3 − I2 ) + Z4 I3 = −1

π 3

(3.13)

I1 V1 1Ω R6

1H L1

cos(t+ π3 )V

+ 1H L6

C3

∼

0.1F

(a)

2Ω R2

R5 1Ω R4 3Ω

I2 j2 2+j

j1 L2 1H

0.1F

Z1 Z2 1+j 1 π3 V 1 V2 ∼ Z5 Z6 V3 Z3 −j10

−j30 3−j10 I3

C4

Z4

(b)

Fig. 3.14 (a) A bridge circuit, with the voltage source in the middle; (b) its frequency-domain version

3.2 AC Circuit Analysis

99

Simplifying, we get, in matrix form, ⎡ ⎢ ⎣

(Z1 + Z3 + Z6 )

−Z1

−Z3

I1

⎡

⎤

⎥⎢ ⎥ ⎢ −Z5 ⎦ ⎣ I2 ⎦ = ⎣ 1

−Z1 (Z1 + Z2 + Z5 ) −Z3

⎤⎡

−Z5 (Z3 + Z4 + Z5 )

0

π 3 −1 π3

I3

⎤ ⎥ ⎦

With Z1 = j 1, ⎡ ⎢ ⎣

Z2 =

1 − j8

j2 , 2 + j1

Z3 = −j 10,

Z4 =

−j 30 , 3 − j 10

0.0000 − j 1.0000

0.0000 + j 10.0000

j 10 −1.0000 + j 0.0000

3.7523 − j 10.8257

−j 1

⎤⎡

Z5 = 1,

I1

⎤

Z6 = 1 + j 1

⎡

0

⎤

⎥⎢ ⎥ ⎢ ⎥ 1.4000 + j 1.8000 −1.0000 + j 0.0000 ⎦ ⎣ I2 ⎦ = ⎣ 0.5 + j 0.866 ⎦ −0.5 − j 0.866

I3

The determinant of the impedance matrix is 80.2771 − j 29.1431. Solving the equilibrium equations, we get ⎡

I1

⎤

⎡

1 − j8

⎢ ⎥ ⎢ ⎣ I2 ⎦ = ⎣ I3

0.0000 − j 1.0000

0.0000 + j 10.0000

j 10 −1.0000 + j 0.0000

3.7523 − j 10.8257

−j 1

⎡

⎤−1 ⎡

⎥ 1.4000 + j 1.8000 −1.0000 + j 0.0000 ⎦

0

⎤

⎥ ⎢ ⎣ 0.5 + j 0.866 ⎦ −0.5 − j 0.866

0.2949 + j 0.0024 0.1441 − j 0.0255 0.2501 − j 0.0712

⎤⎡

0

⎤

⎥⎢ ⎢ ⎥ = ⎣ 0.1441 − j 0.0255 0.3519 − j 0.3810 0.1530 − j 0.0441 ⎦ ⎣ 0.5 + j 0.866 ⎦ 0.2501 − j 0.0712 0.1530 − j 0.0441 0.2225 − j 0.0363 −0.5 − j 0.866 ⎡

−0.0925 − j 0.0689

⎤

⎥ ⎢ = ⎣ 0.3912 + j 0.0038 ⎦ −0.0280 − j 0.0640 {IZ1 = −0.4837 − j 0.0727, IZ3 = −0.0645 − j 0.0049, IZ5 = −0.4192 − j 0.0678} Nodal Analysis (V1 − V2 ) V1 (V1 − V3 ) + + =0 Z1 Z6 Z2

(3.14)

((V2 − 1 π3 ) − V3 ) (V2 − V1 ) V2 + + =0 Z1 Z3 Z5

(3.15)

V3 (V3 − V1 ) ((V3 + 1 π3 ) − V2 ) + + =0 Z4 Z2 Z5

(3.16)

(Z2 Z6 + Z1 Z2 + Z1 Z6 )V1 − Z2 Z6 V2 − Z1 Z6 V3 = 0

(3.17)

π −Z3 Z5 V1 + (Z3 Z5 + Z1 Z5 + Z1 Z3 )V2 − Z1 Z3 V3 = Z1 Z3 1 3 −Z4 Z5 V1 − Z2 Z4 V2 + (Z2 Z5 + Z4 Z5 + Z4 Z2 )V3 = −Z2 Z4 1

(3.18) π . 3

(3.19)

100

3 AC Circuits

The determinant of the admittance matrix is −52.5644 + j 55.4253. Solving the equilibrium equations, we get ⎡

V1

⎤ ⎡

−2.2000 + j 2.6000

0.4000 − j 1.2000 1.0000 − j 1.0000

⎤−1 ⎡

⎥ ⎢ ⎥ ⎢ −10.0000 ⎦ ⎣ V2 ⎦=⎣ 0.0000 + j 10.0000 10.0000 − j 9.0000 V3 −2.7523 + j 0.8257 −1.7615 − j 1.8716 4.9138 + j 1.8459 ⎡

0

⎥ ⎢ ⎣ 5.0000+j 8.6603 ⎦ 0.7401−j 2.4613

⎤⎡

−0.8561 − j 0.0565 0.1183 + j 0.0287 0.3395 − j 0.2318

⎤

0

⎤

⎥⎢ ⎢ ⎥ = ⎣ −0.9594 + j 0.0803 0.1721 + j 0.0837 0.4501 − j 0.2103 ⎦ ⎣ 5.0000 + j 8.6603 ⎦ −0.8306 + j 0.0876 0.1186 + j 0.0472 0.4925 − j 0.2759

0.7401 − j 2.4613

⎤

⎡

0.0236 + j 0.1614 ⎥ ⎢ = ⎣ −0.0491 + j 0.6452 ⎦ −0.1299 − j 0.1531

Circuit 3.3 Consider the nodal analysis of the circuit shown in Fig. 3.15a. The corresponding frequency-domain circuit is shown in Fig. 3.15b. The voltage source is in the middle of the bridge. Usually, two equations using KCL is set up at the two ends of the supernode. Let the current through the source be i flowing from right to left. At the left side node, we get (V2 − V1 ) V2 + = i. Z1 Z3

(3.20)

At the right side node, we get ((V2 − 1 π3 ) − V1 ) (V2 − 1 +i+ Z2 Z4

π 3)

= 0.

(3.21)

Since the currents are flowing in the opposite directions, their sum must be equal to zero. ((V2 − 1 π3 ) − V1 ) (V2 − V1 ) (V2 − 1 V2 + + + Z2 Z1 Z3 Z4

π 3)

= 0.

I1 V1 1Ω R5

1H L1

cos(t+ π3 )V

2Ω R2

C3

0.1F

(a)

R4 3Ω

j2 2+j

Z1 1+j 1 V2 Z5

+∼

1H L5

I2

j1 L2 1H

0.1F

(3.22)

Z2 π 3V

∼

Z3 −j10

−j30 3−j10 I3

C4

V3 Z4

(b)

Fig. 3.15 (a) A bridge circuit with the voltage source in the middle; (b) its frequency-domain version

3.2 AC Circuit Analysis

101

Eliminating i and simplifying, we get π −(Z1 Z3 Z4 +Z2 Z3 Z4 )V1 +(Z1 Z3 Z4 +Z2 Z3 Z4 +Z1 Z2 Z4 +Z1 Z2 Z3 )V2 = (Z1 Z2 Z3 +Z1 Z3 Z4 )1 3 (3.23) Applying at KCL at node 1, we get V1 (V1 − V2 ) (V1 − (V2 − 1 + + Z5 Z1 Z2

π 3 ))

= 0.

π (3.24) 3 Note that the current in the source is flowing towards node 2. Since there is no other source in the circuit, this current gets split up at node 1. With (Z1 Z2 + Z2 Z5 + Z1 Z5 )V1 − (Z2 Z5 + Z1 Z5 ))V2 = (−Z1 Z5 )1

Z1 = j 1,

Z2 =

j2 , 2 + j1

Z3 = −j 10,

Z4 =

−j 30 , 3 − j 10

Z5 = 1 + j 1

solving for V1 and V2 , we get 

−46.2385 + j 25.8716 48.3670 − j 16.1101 −2.2000 + j 2.6000

1.4000 − j 2.2000

  .

V1



 =

V2

15.9839 + j 27.1712



1.3660 + j 0.3660

The determinant of the admittance matrix is 56.7046 − j 23.2514, which is nonzero. This is one of the checks on the problem formulation. Solving the equilibrium equations, we get 

V1 V2



 =

−2.2000 + j 2.6000 

=

1.4000 − j 2.2000

0.0348 − j 0.0245 −0.8299 − j 0.0562 0.0493 − j 0.0256 −0.8582 + j 0.1043

 =

−46.2385 + j 25.8716 48.3670 − j 16.1101

0.1094 + j 0.1714 0.2741 + j 0.7585



−1 

15.9839 + j 27.1712



1.3660 + j 0.3660

15.9839 + j 27.1712



1.3660 + j 0.3660

 .

Substituting for V1 and V2 , from either of the previous equations set up at the ends of the source, we get the current through the voltage source as 0.5112 − j 0.1373. Since we have determined the two independent voltages, there is no need to use this current for the analysis of this circuit. V3 = (0.2741 + j 0.7585) − (0.5 + j 0.866) = −0.2259 − j 0.1075 V. An alternate method to avoid processing a supernode is that we insert an impedance in series with the source, as presented in an earlier example. As its value gets reduced compared with other impedances in the circuit, the circuit becomes more closer with a zero impedance series circuit and the result becomes closer to the exact values. A suitably small series impedance is to be selected. For the example circuit, an impedance of Z = 10−9 + j 10−9 in series with the voltage source yields almost the exact values.

102

3 AC Circuits

Loop Analysis The equilibrium equations are of the same form as those derived for the corresponding DC bridge in Chap. 2 with the source changed. Z1 (I1 − I2 ) + Z3 (I1 − I3 ) + Z5 I1 = 0 −Z1 (I1 − I2 ) + Z2 I2 = 1

(3.25) π 3

−Z3 (I1 − I3 ) + Z4 I3 = −1 ⎡ ⎢ ⎣

(Z1 + Z3 + Z5 )

−Z1

−Z1 (Z1 + Z2 ) −Z3

−Z3

⎤⎡

I1

(3.26) π 3

⎤

(3.27) ⎡

⎥⎢ ⎥ ⎢ 0 ⎦ ⎣ I2 ⎦ = ⎣ 1

0 (Z3 + Z4 )

0

π 3 −1 π3

I3

⎤ ⎥ ⎦.

With −j 30 , Z5 = 1 + j 1 3 − j 10 ⎤ ⎤⎡ ⎤ ⎡ ⎡ I1 0 1.0000 − j 8.0000 0.0000 − j 1.0000 0.0000 + j 10.0000 ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎣ 0.0000 − j 1.0000 0.4000 + j 1.8000 0.0000 + j 0.0000 ⎦ ⎣ I2 ⎦ = ⎣ 0.5 + j 0.866 ⎦ −0.5 − j 0.866 0.0000 + j 10.0000 0.0000 + j 0.0000 2.7523 − j 10.8257 I3 Z1 = j 1,

Z2 =

j2 , 2 + j1

Z3 = −j 10,

Z4 =

The determinant of the impedance matrix is 68.3303 + j 5.1009. Solving the equilibrium equations, we get ⎡

I1

⎤

⎡

1.0000 − j 8.0000 0.0000 − j 1.0000 0.0000 + j 10.0000

⎤−1 ⎡

⎢ ⎥ ⎢ ⎥ ⎣ I2 ⎦ = ⎣ 0.0000 − j 1.0000 0.4000 + j 1.8000 0.0000 + j 0.0000 ⎦ 0.0000 + j 10.0000 0.0000 + j 0.0000 2.7523 − j 10.8257 I3 ⎡

0.3003 − j 0.0133 0.1605 + j 0.0283 0.2576 − j 0.0778

⎤⎡

0

⎤

⎥ ⎢ ⎣ 0.5 + j 0.866 ⎦ −0.5 − j 0.866 0

⎤

⎥⎢ ⎢ ⎥ = ⎣ 0.1605 + j 0.0283 0.1993 − j 0.4955 0.1455 − j 0.0109 ⎦ ⎣ 0.5 + j 0.866 ⎦ 0.2576 − j 0.0778 0.1455 − j 0.0109 0.2284 − j 0.0375 ⎡

−0.1404 − j 0.0310

−0.5 − j 0.866

⎤

⎥ ⎢ = ⎣ 0.4466 − j 0.1958 ⎦ −0.0645 − j 0.0584

Circuit 3.4 Figure 3.16a shows a circuit with a current source. The corresponding tree is shown in Fig. 3.16b. The left voltage source is sin(t) = cos(t − π2 ). Therefore, V1 = −j is known and we have to solve for V2 and V3 only. The equilibrium equations are (V2 − (−j 1)) V2 + = −j 1 Z1 Z2 (V3 + 2) V3 + = j 1. Z4 Z5

(3.28) (3.29)

3.2 AC Circuit Analysis

Z4 = −j10

Z1 j0.1 V1 = −j

+ ∼

(a)

∼ +

2 cos(t)V

2 0 ∼ +

V3

I1

sin(t)V

sin(t)V

V1 −j + ∼

V2

Z5 = 1+j0.2

Z1 j0.1

Z2 = 2−j5

2Ω

sin(t)A ∼ −j V3 I2 I3

2 cos(t)V

I1

Z3

Z4 = −j10

V2

103

(b)

Fig. 3.16 (a) A circuit with a current source; (b) the corresponding tree

Simplifying and solving, we get (Z1 + Z2 )V2 = (−j 1)(Z1 Z2 + Z2 )

(3.30)

(Z4 + Z5 )V3 = j (Z4 Z5 ) − 2Z5

(3.31)

V2 =

(−j 1)(Z1 Z2 + Z2 ) (Z1 + Z2 )

(3.32)

V3 =

j (Z4 Z5 ) − 2Z5 . (Z4 + Z5 )

(3.33)

With Z1 = j 0.1, we get

Z2 = 2 − j 5,

Z3 = 2,

Z4 = −j 10,

Z5 = 1 + j 0.2

{V2 = 0.0946 − j 1.0182, V3 = −0.0791 + j 0.8244}.

Mesh Analysis Applying KVL around the left side loop, we get, with I2 = j 1, I1 Z1 + Z2 (I1 − I2 ) = −j 1 I1 (Z1 + Z2 ) = −j 1 + Z2 j 1 I1 =

j 1(Z2 − 1) = 0.1821 + j 0.9461 Z1 + Z2

V2 = −I1 Z1 − j = 0.0946 − j 1.0182 Applying KVL around the right side loop, we get, −(I2 − I3 )Z4 + Z5 I3 = −2 I3 =

−2 + j Z4 = 0.0824 + j 0.8079 Z4 + Z5

V3 = I3 Z5 = −0.0791 + j 0.8244.

104

3 AC Circuits

An alternative method to avoid processing a supermesh is that we insert an impedance in parallel with the source. As its value gets increased compared with other impedances in the circuit, the circuit becomes more closer with an infinite impedance in parallel and the result becomes closer to the exact values. A suitably high parallel impedance is to be selected. For the example circuit, the values are almost the same as the exact ones, with parallel impedance 100 + j 100.

Circuit 3.5 In the last example, a current source exists only in one loop. Let us consider a circuit with a current source between two loops. The circuit and the tree are shown, respectively, in Fig. 3.17a, b. There are two nodes. But, one node voltage is given. Similarly, there are two loops. But, the currents are related. Therefore, the circuit analysis reduces to one variable problem for both the nodal and loop analyses. With −j 5 , Z3 = 2, V1 = −j V . Z1 = j 0.1, Z2 = 1 − j5 With the current source supplying 2 A towards the ground node, KVL around the circuit and the current source constraint yields Z1 I1 + Z3 I3 = −j 1. Since I1 − I3 = 2 Z1 I1 + Z3 (I1 − 2) = −j 1 or

(Z1 + Z3 )I1 = −j 1 + 2Z3 .

Solving, we get I1 = 1.9701 − j 0.5985 and I3 = I1 − 2 = −0.0299 − j 0.5985. Nodal Analysis V2 (V2 − (−j 1)) + = −2. Z1 Z3

v2

V2

i2

i3 C2 0.2F R2 R3 2Ω 1Ω V1 −j1V ∼

2 cos(t)A

sin(t)V

v1 + ∼

C2 0.2F

R2 1Ω

(a)

+ ∼

∼ (b)

Fig. 3.17 (a) A circuit with a current source between loops; (b) the corresponding tree

2 cos(t)A

i1

0.1H

sin(t)V

L1

3.2 AC Circuit Analysis

105

Solving for V2 , we get V2 =

−2Z1 Z3 − j Z3 = −0.0599 − j 1.1970. Z1 + Z3

Let us do this problem considering one source at a time. When we consider the voltage source alone, the current source has to be open-circuited. Then, by voltage division, we get the voltage across the 2  resistor as −j 2 = −0.0499 − j 0.9975. 2 + j 0.1 When we consider the current source alone, the voltage source has to be short-circuited. Then, by current division, we get the voltage across the 2  resistor as −2

j 0.1(2) = −0.0100 − j 0.1995. 2 + j 0.1

By adding the partial voltages, we get V2 = −0.0599 − j 1.1970 as obtained earlier.

3.2.8

Linearity Property of Circuits

Consider the circuit shown in Fig. 3.18. The circuit has three sources, A DC voltage source with magnitude 2 V with the positive terminal connected to the ground node. An AC voltage source sin(t + π 3 ) V, with magnitude 1 V at frequency 1 rad/s, is placed with the negative terminal connected to the ground node. An AC current source 2 cos(5t) A, with magnitude 2 A at frequency 5 rad/s, is placed with the the current flowing from right to left. Since different frequencies are involved and impedances vary with frequencies, we have to use the linearity property of the circuits and the resulting superposition theorem to analyze this circuit. The total response is the sum of the individual timedomain responses. In the analysis of AC circuits with sources of different frequencies, the use of linearity is required. The transform analysis of circuits is based on linearity. For DC circuits, linearity may provide simplified analysis for some circuits.

2 cos(5t)A V3 ∼ I I3 2 2Ω Z2

V1 1 − π6 + ∼ Z

3

1 1+jω0.2

10

+ 2V

Z6=jω0.2

I1 Z1 jω0.1

2 Z5= 1+j2ω0.1

Z4

V2

sin(t+π3 )V

Fig. 3.18 A circuit with voltage and current sources of different frequencies

Z7 1

106

3 AC Circuits

Table 3.2 Voltages due to each source and their total in circuit shown in Fig. 3.18 Source

V1

V2

V3

V source, ω = 1 DC, ω = 0 I source, ω = 5 Total

1 (− π6 ) 0 0 cos(t − π6 )

0.8615 − j 0.5079 0 0.0476 + j 1 cos(t − 0.5327) + 1.0011 cos(5t + 1.5232)

0 − 32 −2 −2 cos(5t) −

2 3

Table 3.3 Currents due to each source and their total in circuit shown in Fig. 3.18 Source

I1

I2

I3

V source DC I source Total

0.0794 − j 0.0449 0 −(2.0000 − j 0.0952) 0.0912 cos(t − 0.5152) − 2.0023 cos(5t − 0.0476)

0 0 −2 −2 cos(5t)

0 − 23 (−1 + j 1) √ 2 cos(5t +

3π 2 4 )− 3

Consider the effect of the DC source alone. Since the frequency of DC source is zero, the reactance due to an inductance is zero and that due to a capacitance is infinity. Therefore, an inductance, in effect, is a short-circuit and a capacitance is an open-circuit. Further, an ideal current source is replaced by an open-circuit and an ideal voltage source is replaced by a short-circuit. Consequently, the circuit reduces to the DC source with amplitude −2 V and the two resistors, with values 2  in Z5 and 1  in Z7 , connected in series. Therefore, I3 = − 23 A and V3 = − 23 V. All other voltages and currents are zero, as shown in Tables 3.2 and 3.3. With Z1 = j ω0.1, Z2 =

1 , 1 + j ω0.2

Z3 = 10, Z4 = 2, Z5 =

2 , Z6 = j ω0.2, Z7 = 1, 1 + j 2ω0.1

let us find the response of the circuit to the voltage source sin(t +

π π ) = cos(t − ) 3 6

alone. Impedance Z2 is due to the parallel connection of a resistor with 1  and a capacitor with C = 0.2 F. With ω = 1, Z1 = j 0.1,

Z2 =

1 ) (1)( j 0.2

1+

1 j 0.2

= 0.9615 − j 0.1923,

Z3 = 10, V = 1 −

π = 0.8660 − j 0.5 6

Impedance Z5 is due to the parallel connection of a resistor with 2  and a capacitor with C = 0.1 F. The right side of the circuit gets disconnected as the current source is replaced by an open-circuit. The DC source is replaced by a short-circuit. Therefore, V3 = 0, I2 = 0, and I3 = 0. V1 = 1 (− π6 ). I1 =

V1 = 0.0794 − j 0.0449 Z1 + Z2 + Z3

V2 = I1 (Z2 + Z3 ) = V1 − I1 Z1 = 0.8615 − j 0.5079.

3.2 AC Circuit Analysis

107

With ω = 5, Z1 = j 0.5,

Z2 = 0.5 − j 0.5, Z3 = 10, Z4 = 2, Z5 = 1 − j 1,

Z6 = j 1,

Z7 = 1

Let us find the response of the circuit with the current source 2 cos(5t) alone. Note that the frequency ω is 5 radians, not 1 compared with the AC voltage source. Therefore, the impedance values are different. The two voltages sources are replaced by short-circuits. Applying KCL at nodes 3 and 2 V3 V3 + +2=0 Z5 Z6 + Z7 V3 =

−2(Z5 (Z6 + Z7 )) = −2 Z5 + Z6 + Z7

V2 V2 + −2=0 Z1 Z2 + Z3 V2 = −I1 =

2(Z1 (Z2 + Z3 )) = 0.0476 + j 1 Z1 + Z2 + Z3

V2 = 2 − j 0.0952, Z1

I2 = −2,

I3 =

V3 = −1 + j 1. Z6 + Z7

Using current-division formula also, we can find the currents I1 and I3 . Voltages and currents due to each source and their totals are shown in Tables 3.2 and 3.3.

Circuit 3.6 Consider the circuit with a voltage-controlled voltage source, shown in Fig. 3.19. Voltage source is 2 cos(t) V. Current source is sin(t) = cos(t − π2 ) A. In the frequency-domain, the voltage source is 2 and the current source is −j 1 = 1 (− π2 ). The circuit contains a controlled voltage source. Therefore, although there are five nodes, three equilibrium equations are enough. With the angular frequency of the sources being 1 radian,

Fig. 3.19 A circuit with a voltage-controlled voltage source

Z2 = 1,

Z3 = 3 − j 10,

i1 v1

+ −

Z1 = 2 + j 0.1,

2(v1 −v4 )

R1 2Ω

L1 0.1H

Z4 = 3,

Z5 = 2.

R5 2Ω v2 R 2 1Ω ∼ sin(t)A

2 cos(t)V v3 + ∼ v4 i2 i3

C3

R3

0.1F

3Ω

R4

3Ω

108

3 AC Circuits

Loop Analysis The equilibrium equations are (Z5 I1 + Z4 I3 − 2Z5 I1 ) + (Z2 (I1 − I2 − I3 )) − Z3 I2 = 0 −Z1 (I (2) + I (3) + j 1) = Z5 I1 + Z4 I3 Z3 I2 − Z4 I3 = 2. The first equation corresponds to the middle loop involving resistors R2 and R3 . To write the KVL equation around the loop, we need the voltage V2 . V1 = Z5 I1 + Z4 I3

and

V1 − V4 = Z5 I1 .

Therefore, the voltage across the controlled voltage source is 2Z5 I1 , with opposite polarity to that of V1 . Consequently, V2 = (Z5 I1 + Z4 I3 − 2Z5 I1 ). The second equation corresponds to the leftmost loop involving resistor Z1 . The current entering Z1 is (I (2) + I (3) − (−j 1) from the ground side. Therefore, we get another expression for V1 involving the current source. V1 = −Z1 (I (2) + I (3) + j 1) Equating these two equations yields an equilibrium equation. The last equation corresponds to the rightmost loop involving impedances Z3 and Z4 . Rearranging the equations, we get (−Z5 + Z2 )I1 − (Z2 + Z3 )I2 + (Z4 − Z2 )I3 = 0 Z5 I1 + Z1 I2 + (Z1 + Z4 )I3 = −j Z1 0I1 + Z3 I2 − Z4 I3 = 2. Substituting the numerical values for the impedances and using matrices, we get ⎡ ⎢ ⎣

−1 −4 + j 10 2 0

2

⎤⎡

I1

⎤

⎡

0

⎤

⎥ ⎥⎢ ⎥ ⎢ 2 5 + j 0.1 ⎦ ⎣ I2 ⎦ = ⎣ −j Z1 ⎦ 2 I3 3 − j 10 −3

The determinant of the impedance matrix is nonzero, 10 − j 29.4. The determinant of the impedance matrix must be nonzero. Otherwise, the equations are not independent. Verify that the product of the impedance matrix and its inverse is the identity matrix of the same size. One of the equations must be based on the relation between dependent and independent source. Solving for the currents, we get ⎡

I1

⎤

⎡

⎢ ⎥ ⎢ ⎣ I2 ⎦ = ⎣ I3

−1 −4 + j 10 2 0

2

⎤−1 ⎡

⎥ 2 5 + j 0.1 ⎦

3 − j 10

−3

⎤ 0 ⎣ 0.1 − j 2 ⎦ 2

3.2 AC Circuit Analysis

⎡ ⎢ =⎣ ⎡

109

−1.7342 − j 0.1584 −0.3671 − j 0.0792 −1.7653 − j 0.2499 0.0622 + j 0.1829

0.0311 + j 0.0915

0.6719 − j 0.0245

0.3360 − j 0.0122

−3.7257 + j 0.2264

⎤⎡

0

⎤

⎥ ⎥⎢ 0.0903 + j 0.2754 ⎦ ⎣ 0.1 − j 2 ⎦ 2 0.6750 − j 0.0255

⎤

⎢ ⎥ = ⎣ 0.3666 + j 0.4978 ⎦ 1.3591 − j 0.7242 Let us find the node voltages and verify their relations. V3 = I2 Z3 = (0.3666 + j 0.4978)(3 − j 10) = 6.0774 − j 2.1726 V4 = I3 Z4 = (1.3591 − j 0.7242)(3) = 4.0774 − j 2.1726 V1 = V4 + I1 Z5 = (4.0774 − j 2.1726) + (−3.7257 + j 0.2264)(2) = −3.3741 − j 1.7197 V2 = 2V4 − V1 = 2(4.0774 − j 2.1726) − (−3.3741 − j 1.7197) = 11.5288 − j 2.6254 Now, V3 − V4 = 2 V1 − V2 = (−3.3741 − j 1.7197) − (11.5288 − j 2.6254) = −14.9028 + j 0.9057 = 2I1 Z1 = 2(−3.7257 + j 0.2264)(2 + j 0.1) V3 − V2 = −5.4514 + 0.4529 = (I1 − I2 − I3 )Z2

The current through the independent voltage source flowing from left to right is ir = I1 − I3 = −5.0848 + j 0.9506. The current through the dependent voltage source flowing from left to right is il = I1 − I2 − I3 − j = −5.4514 + j 0.4529 − j = −5.4514 − j 0.5471. Nodal Analysis The voltage source is not connected to the ground node. Therefore, we have to find the current through it. Setting up two KCL equations at the terminals of the source, we get (V4 − V1 ) V4 (V2 − V3 ) V3 + − + = 0. Z5 Z4 Z2 Z3 The dependent voltage source is also not connected to the ground node. Therefore, we have to find the current through it. Setting up two KCL equations at the terminals of the source, we get (V3 − V2 ) (V1 − V4 ) V1 + (−j 1) − − = 0. Z2 Z5 Z1

110

3 AC Circuits

Now, applying KVL to the branches involving the dependent voltage source, we get the equilibrium equations as (V3 − V2 ) (V1 − V4 ) V1 + (−j 1) − − =0 Z2 Z5 Z1

(3.34)

(V4 − V1 ) V4 (V2 − V3 ) V3 + − + =0 Z5 Z4 Z2 Z3

(3.35)

V1 + V2 − 2V4 = 0.

(3.36)

Replacing

V3 = (V4 + 2),

we get ((V4 + 2) − V2 ) (V1 − V4 ) V1 + (−j 1) − − =0 Z2 Z5 Z1

(3.37)

(V4 − V1 ) V4 (V2 − (V4 + 2)) (V4 + 2) + − + =0 Z5 Z4 Z2 Z3

(3.38)

V1 + V2 − 2V4 = 0

(3.39)

Simplifying, we get −(Z1 Z2 + Z2 Z5 )V1 − Z1 Z5 V2 + (Z1 Z2 + Z1 Z5 )V4 = −2Z1 Z5 + j Z1 Z2 Z5 −Z2 Z3 Z4 V1 − Z3 Z4 Z5 V2 + (Z2 Z3 Z4 + Z2 Z3 Z5 + Z3 Z4 Z5 + Z2 Z4 Z5 )V4 = −2(Z2 Z4 Z5 + Z3 Z4 Z5 ) V1 + V2 − 2V4 = 0

With Z1 = 2 + j 0.1,

Z4 = 3, Z5 = 2 ⎤⎡ ⎤ ⎡ ⎤ V1 −0.0400 − j 0.0010 −0.0400 − j 0.0020 0.0600 + j 0.0030 −0.082 + j 0.036 ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎣ −0.0900 + j 0.3000 −0.1800 + j 0.6000 0.3900 − j 1.1000 ⎦ ⎣ V2 ⎦ = ⎣ −0.480 + j 1.200 ⎦ 0 0.0100 + j 0.0000 0.0100 + j 0.0000 −0.0200 + j 0.0000 V4 ⎡

Z2 = 1,

Z3 = 3 − j 10,

The determinant of the admittance matrix is −20 + j 58.8. ⎡

⎤

⎡

⎤−1 ⎡

⎤ −0.082 + j 0.036 ⎥ ⎣ ⎢ ⎥ ⎢ ⎣ V2 ⎦ = ⎣ −0.090 + j 0.300 −0.180 + j 0.600 0.390 − j 1.100 ⎦ −0.480 + j 1.200 ⎦ 0 0.010 + j 0.000 0.010 + j 0.000 −0.020 + j 0.000 V4 ⎤ ⎡ ⎤⎡ −0.1369 + j 0.0976 0.0088 + j 0.0310 1.4525 + j 0.3903 −0.082 + j 0.036 ⎥ ⎢ ⎥⎢ = 100 ⎣ −0.8710 − j 0.0609 −0.0119 − j 0.0300 −4.4842 − j 0.2435 ⎦ ⎣ −0.480 + j 1.200 ⎦ −0.5040 + j 0.0184 −0.0015 + j 0.0005 −2.0158 + 0.0734 0 ⎤ ⎡ −3.3741 − j 1.7197 ⎥ ⎢ = ⎣ 11.5288 − j 2.6254 ⎦ 4.0774 − j 2.1726 V1

−0.040 − j 0.001 −0.0400 − j 0.002

0.060 + j 0.003

The voltages are the same as those obtained by the loop method.

3.2 AC Circuit Analysis

111

Nodal or Mesh Analysis Basically, both types of circuit analysis are important, since one type may be conceptually and/or computationally simpler for a given circuit. In certain applications, one of the two methods is preferred or applicable. Further, the solution obtained by one method can be checked using the other method. Therefore, it is good to be familiar with both the methods. In nodal analysis, if the circuit has a voltage source connected to the ground node, then that voltage can be considered as one independent variable and the number of equilibrium equations can be reduced. Further, if a current source, without a parallel resistance, is connected between non-ground nodes, then nodal method is advantageous. Similarly, if a voltage source is connected between two non-ground nodes, then mesh analysis is easier. The solution to the last example is simpler by the mesh method. Further, if the circuit has a current source connected to the ground node, then that current can be considered as one independent variable and the number of equilibrium equations can be reduced.

Circuit 3.7 A circuit with a current-controlled current source is shown in Fig. 3.20. Nodal Analysis Since V3 = j V is known, there are only 2 unknowns, {V1 , V2 }. When a voltage source is connected between a reference node and a nonreference node, the problem is simplified. The equilibrium equations are (V2 − V1 ) (V2 − j ) (V2 ) + + =0 Z2 Z4 Z3

(3.40)

V1 (V1 − j ) V1 (V1 − V2 ) + −2 + = 0. Z1 Z5 Z1 Z2

(3.41)

The first equation is due to the application of KCL at node with voltage V2 . The second equation is due to the application of KCL at node with voltage V1 . Simplifying, we get

Fig. 3.20 A circuit with a current-controlled current source

− Z3 Z4 V1 + (Z3 Z4 + Z2 Z4 + Z2 Z3 )V2 = j Z2 Z3

(3.42)

(−Z2 Z5 + Z1 Z5 + Z1 Z2 )V1 − Z1 Z5 V2 = j Z1 Z2 .

(3.43)

i1

v1

R5 1Ω R2

ix

3Ω

L1 0.1H

R1 1Ω

2ix

v2 i2

L3 0.1H

R4

v3

3Ω

i3

R3 1Ω

sin(t)V ∼ +

112

3 AC Circuits

With Z1 = 1 + j 0.1,

Z2 = 3,

Z3 =

j 0.1 , 1 + j 0.1

Z4 = 3,

Z5 = 1,

we get, in matrix form, 

−0.0297 − j 0.2970

9.0594 + j 0.5941



1.0000 + j 0.4000 −1.0000 − j 0.1000

V1



V2

 =

−0.2970 + j 0.0297



−0.3 + j 3

.

The determinant of the admittance matrix is −8.8218 − j 3.9178. Solving for the voltages, we get 

V1 V2



 =

9.0594 + j 0.5941

−1 

−0.2970 + j 0.0297

1.0000 + j 0.4000 −1.0000 − j 0.1000 

=

0.0989 − j 0.0326 0.8827 − j 0.3247



0.6808 + j 2.7582



−0.3 + j 3 −0.2970 + j 0.0297

0.1115 − j 0.0042 0.0153 + j 0.0269 

=

−0.0297 − j 0.2970



−0.3 + j 3



−0.1182 + j 0.0424

I1 =

(V1 − j ) = 0.6808 + j 1.7582 Z5

I2 =

V2 = 0.3058 + j 1.2245 Z3

I3 = I1 +

(V2 − j ) = 0.6414 + j 1.4390. Z4

Mesh Analysis IZ1 = I2 + I3 Z1 (I2 + I3 ) + Z2 (I1 − I2 − I3 ) − Z3 I2 = 0 Z3 I2 + Z4 (I1 − I3 ) = j Z1 (I2 + I3 ) − Z5 I1 = j. The first equation is obtained by equating the voltages on both sides of the current source. The second equation is obtained by applying KVL to the rightmost loop involving the voltage source. The third equation is obtained by applying KVL to the loop involving the voltage source and impedances Z1 and Z5 . Voltage source is sin(t) V. Simplifying, we get Z2 I1 + (Z1 − Z2 − Z3 )I2 + (Z1 − Z2 )I3 = 0 Z4 I1 + Z3 I2 − Z4 I3 = j −Z5 I1 + Z1 I2 + Z1 I3 = j.

3.2 AC Circuit Analysis

113

With Z1 = 1 + j 0.1, ⎡

Z2 = 3,

Z3 =

j 0.1 , 1 + j 0.1

Z4 = 3,

3.0000 + j 0.0000 −1.0000 + j 0.0000

⎤⎡

⎡ ⎤ 0 ⎥⎢ ⎥ ⎢ ⎥ 0.0099 + j 0.0990 −3.0000 + j 0.0000 ⎦ ⎣ I2 ⎦ = ⎣ j ⎦ . j 1.0000 + 0.1000 1.0000 + j 0.1000 I3

3.0000 + j 0.0000 −2.0099 + j 0.0010 −2.0000 + j 0.1000

⎢ ⎣

Z5 = 1

I1

⎤

The determinant of the impedance matrix is nonzero, 2.9406 + j 1.3059. ⎡

I1

⎤

⎡

⎤−1 ⎡ ⎤ 0 ⎥ ⎢ ⎥ 0.0099 + j 0.0990 −3.0000 + j 0.0000 ⎦ ⎣ j ⎦ j 1.0000 + 0.1000 1.0000 + j 0.1000

3.0000 + j 0.0000 −2.0099 + j 0.0010 −2.0000 + j 0.1000

⎢ ⎥ ⎢ 3.0000 + j 0.0000 ⎣ I2 ⎦ = ⎣ −1.0000 + j 0.0000 I3

⎤⎡ ⎤ 0 ⎥⎢ ⎥ ⎢ = ⎣ −0.0378 − j 0.0852 0.3345 − j 0.0125 0.8900 − j 0.2932 ⎦ ⎣ j ⎦ j 0.9053 − 0.2664 −0.3192 + j 0.0394 1.7582 − j 0.6808 ⎡

⎡

0.9026 − j 0.2648

0.6808 + j 1.7582

0.0126 + j 0.0284 1.7456 − j 0.7093

⎤

⎥ ⎢ = ⎣ 0.3058 + j 1.2245 ⎦ . 0.6414 + j 1.4390

Circuits with One or Two Variables Now, we are going to present the analysis of simpler circuits with one or two variables, which can be solved manually. Circuit in Fig. 3.21 The circuit shown in Fig. 3.21 has two voltage sources. Nodal Analysis Z1 = 2,

Z2 = j 2,

Z3 = 3,

V1 = 1,

V3 = 2.

Voltage V2 is the only unknown. Therefore, applying KCL at node 2, we get Fig. 3.21 A circuit with two voltage sources

R1 i1

2Ω

v1 + cos(2t)V ∼

v2 i2 L2

R3 3Ω i3 v3 1H + ∼ 2 cos(2t)V

114

3 AC Circuits

(V2 − 1) (V2 − 2) V2 + + = 0. Z1 Z3 Z2 Solving, we get V2 =

Z2 Z3 + 2Z1 Z2 = (1.0294 + j 0.6176) V. Z1 Z2 + Z2 Z3 + Z1 Z3

The currents are I1 = −

(V2 − 1) (V2 ) = (−0.0147 − j 0.3088) A, I2 = = (0.3088 − j 0.5147) A, Z1 Z2 I3 =

(V2 − 2) = (−0.3235 + j 0.2059) A. Z3

With −I1 + I2 + I3 = 0, KCL is satisfied. Mesh Analysis I2 = I1 − I3 . Z1 I1 + Z2 (I1 − I3 ) = 1 Z3 I3 − Z2 (I1 − I3 ) = −2. Simplifying, we get (Z1 + Z2 )I1 − Z2 I3 = 1 −Z2 I1 + (Z2 + Z3 )I3 = −2. Substituting the numerical values, we get the impedance matrix Z as

Z=

 2 + j 2 −j 2 . −j 2 3 + j 2

Solving the equations, we get the same values of currents, as found earlier. Let us do the problem by superposition method. Let us find the response to the source at the left side. Then, we replace other source by a short-circuit. Current I1 is I1 =

1 = (0.2794 − j 0.1324). Z1 + (Z2 ||Z3 )

By current division, we get I2 = I1

Z3 = (0.1324 − j 0.2206), I3 = I1 − I2 = (0.1471 + j 0.0882). Z2 + Z3

Let us find the response to the response to the source at the right side. Then, we replace the other source by a short-circuit. Current I3 is I3 = − By current division, we get

2 = (−0.4706 + j 0.1176). Z3 + (Z2 ||Z1 )

3.2 AC Circuit Analysis

I1 = I3

115

Z2 = (−0.2941 − j 0.1765), I2 = I1 − I3 = (0.1765 − j 0.2941). Z2 + Z1

Adding the corresponding two responses, we get the same values for the currents. Let us find the current I2 through Z2 using The´venin’s theorem. The equivalent impedance, with Z2 disconnected, is 6 Zeq = Z1  Z2 = 2  3 = . 5 Due to the 1 V source alone, V2 = 3/5 V. Due to the 2 V source alone, V2 = 4/5 V. Therefore, V2 = Voc = 7/5. The current through Z2 is I2 =

7 Voc 7/5 = = 0.3088 − j 0.5147 = Z2 + Zeq 6/5 + j 2 6 + j 10

as found earlier.

Circuit in Fig. 3.22 Figure 3.22 shows a circuit with voltage and current sources. Nodal Analysis Z1 = 2,

Z2 = j 2,

Z3 = 3,

I1 = −1,

V3 = 2.

Voltage V2 is the only unknown. Therefore, applying KCL at node 2, we get (V2 − 2) (V2 ) + = −1. Z3 Z2 Solving, we get V2 =

−Z2 Z3 + 2Z2 = (−0.3077 − j 0.4615) V. Z2 + Z3

The currents are I1 = −

(V2 − 1) (V2 ) = (−0.0147 − j 0.3088)A, I2 = = (0.3088 − j 0.5147)A, Z1 Z2

(V2 − 2) = (−0.3235 + j 0.2059)A Z3 With −I1 + I2 + I3 = 0, KCL is satisfied. I3 =

Fig. 3.22 A circuit with voltage and current sources

R1 i1

2Ω

v2 i2 L2

∼

cos(2t)A

R3 3Ω i3 v3 1H + ∼ 2 cos(2t)V

116

3 AC Circuits

Mesh Analysis I2 = I1 − I3 , I1 = −1. Z3 I3 − Z2 (I1 − I3 ) = −2. Solving, we get I3 =

−2 + I1 Z2 , Z3 + Z2

I3 = (−0.7692 − j 0.1538) A.

We get the same values of currents, as found earlier. Let us solve the problem by superposition method. Let us find the response due to the voltage source. The current source is open-circuited. Then, I3 = −

2 = (−0.4615 + j 0.3077), I2 = −I3 . Z2 + Z3

Let us find the response due to the current source. The voltage source is short-circuited. Then, by current division, I3 = −

Z2 Z3 = (−0.3077 − j 0.4615), I2 = − = (−0.6923 + j 0.4615) Z2 + Z3 Z2 + Z3

Adding the corresponding two responses, we get the same values for the currents. Let us find the current I3 through Z3 using The´venin’s theorem. The equivalent impedance, with Z3 disconnected, is Zeq = Z2 = j 2. Remember that the current source is open-circuited. Due to the 1 A source, V2 = −j 2V . Then, Voc = −j 2 − 2. Therefore, the current through Z3 is I3 =

Voc −j 2 − 2 = −0.7692 − j 0.1538 = Z3 + Zeq 3 + j2

as found earlier.

Circuit in Fig. 3.23 Figure 3.23 shows a circuit with voltage and current sources. Z1 = 2,

Z2 = j 2,

Z3 = 3,

I3 = 1,

V1 = 2.

Mesh Analysis I1 − I2 = I3 I1 Z1 + I2 Z2 = I1 Z1 + (I1 − I3 )Z2 = 2,

I1 =

With −I1 + I2 + I3 = 0, KCL is satisfied. V2 = I2 Z2 = 0.

2 + I3 Z2 = 1, Z2 + Z1

I2 = 0.

3.2 AC Circuit Analysis

117

Fig. 3.23 A circuit with voltage and current sources

R1 i1

2Ω

v1

v2 i2

R3 3Ω i3

L2 1H

+ 2 cos(2t)V ∼

∼

cos(2t)A

Nodal Analysis V2 (V2 − 2) + = −1. Z2 Z1 Solving, V2 = 0 and VR3 = 3V . The voltage across the current source is −3 V. Let us do the problem by superposition method. Consider the response due to the voltage source alone. The current source is open-circuited. Then, I1 = I2 =

2 . 2 + j2

Consider the response due to the current source alone. The voltage source is short-circuited. Then, by current division, we get j2 2 I1 = , I2 = I1 − I3 = − . 2 + j2 2 + j2 Adding the partial currents, we get the same results. Let us find the current I2 through Z2 using The´venin’s theorem. The equivalent impedance, with Z2 disconnected, is Zeq = Z1 = 2. Remember that the current source is open-circuited. Due to the 1 A source alone, V2 = − Due to the 2 V source alone, V2 =

j2 × 2 . 2 + j2

j2 × 2 . 2 + j2

Then, Voc = 0 and I2 = 0, as found earlier.

Circuit in Fig. 3.24 Figure 3.24 shows a circuit with two current sources. Z1 = 2,

Z2 = j 2,

Z3 = 3,

I3 = 1,

I1 = −2.

118

3 AC Circuits

Fig. 3.24 A circuit with two current sources

R1 i1

2Ω

v2 i2

R3 3Ω i3

L2 1H ∼

2 cos(2t)A

∼

cos(2t)A

Nodal Analysis V2 = −3 and V2 = −j 6V Z2 I1 = −2, I2 = −

j6 = −3A, I3 = 1 A. j2

With −I1 + I2 + I3 = 0, KCL is satisfied. Mesh Analysis I3 = 1,

I1 = −2,

I2 = (I1 − I3 ) = −3A.

Let us do the problem by superposition method. Consider the response due to the 2 A current source alone. The other current source is open-circuited. Then, by current division, we get I1 = −2,

I2 = −2,

I3 = 0.

Consider the response due to the 1 A current source alone. The other current source is open-circuited. Then, by current division, we get I1 = 0,

I2 = −1,

I3 = 1.

Adding the two partial results, we get the same currents.

Circuit in Fig. 3.25 A circuit with a voltage-controlled voltage source is shown in Fig. 3.25. Z1 = 2,

Z2 = j 1,

Z3 = 3

By nodal analysis, we get (V2 − 1) V2 V2 − (V2 − 1) + + = 0 or Z1 Z2 Z3

V2 =

Z3 Z2 − Z1 Z2 = (0.0667 + j 0.1333) Z1 Z3 + Z2 Z3

The currents are I1 = −

(V2 − 1) = (0.4667−j 0.0667)A, Z1

With −I1 + I2 + I3 = 0, KCL is satisfied.

I2 =

V2 = (0.1333−j 0.0667)A, Z2

I3 = I1 −I2 =

1 A 3

3.2 AC Circuit Analysis

119

Fig. 3.25 A circuit with a voltage-controlled voltage source

v2 i1 R1

i2 2Ω

v1 + L2 cos(t)V ∼

R3 i3 3Ω + − (v2 −v1) 1H

Mesh Analysis Z1 (I2 + I3 ) + I2 Z2 = 1,

or

(Z2 + Z1 )I2 + Z1 I3 = 1

Z3 I3 − Z2 I2 = Z1 (I3 + I2 ),

− (Z2 + Z1 )I2 + (Z3 − Z1 )I3 = 0.

or

Solving, we get the same currents obtained earlier. Let us find the current I2 through Z2 using The´venin’s theorem. (V2 − 1) V2 − (V2 − 1) + =0 Z1 Z3

or

V2 = Voc =

Z3 − Z1 1 = . Z3 3

Let us apply a 1 A current source instead of the load impedance and short-circuit the independent voltage source. Then, Zeq = V2 = Z1 = 2. Therefore, the current through Z2 is I2 =

Voc 1/3 = 0.1333 − j 0.0667 = Z2 + Zeq 2 + j1

as found earlier.

Circuit in Fig. 3.26 A circuit with a current-controlled voltage source is shown in Fig. 3.26. Z1 = 2, By nodal analysis, we get

Z2 = j 1,

Z3 = 3.

(V2 − 1) V2 V2 + 2j V2 + + = 0. Z1 Z2 Z3

Solving, we get V2 =

Z3 Z2 = (0.5172 + j 0.2069) V. Z3 Z2 + Z3 Z1 + Z1 Z2 + j 2Z1 Z2

120

3 AC Circuits

Fig. 3.26 A circuit with a current-controlled voltage source

i1 R1

v2

R3

i2

3Ω

2Ω

v1 + L2 cos(t)V ∼

i3 + − 2i2

1H

The currents are I1 = (0.2414 − j 0.1034)A,

I2 = (0.2069 − j 0.5172)A,

I3 = (0.0345 + j 0.4138)A

With −I1 + I2 + I3 = 0, KCL is satisfied. Mesh Analysis Z1 I1 + I2 Z2 = 1 (I1 − I2 )Z3 − I2 Z2 + 2I2 = 0 or Z3 I1 + I2 (2 − Z3 − Z2 ) = 0 Solving, we get the same currents obtained earlier. Let us find the current I3 through Z3 using The´venin’s theorem. (V2 − 1) V2 + = 0 or Z1 Z2 Voc = V2 −

V2 =

Z2 j1 = Z1 + Z2 2 + j1

2V2 = −0.6 + j 0.8. Z2

Let us apply a 1 A current source instead of the load impedance and short-circuit the independent voltage source. Then, the voltage across the current source is Zeq . I2 =

2 j2 and V2 = 2 + j1 2 + j1

Zeq = V2 − 2I2 = −1.2 + j 1.6. Therefore, the current through Z3 is I3 = as found earlier.

Voc (−0.6 + j 0.8) = = 0.0345 + j 0.4138 Z3 + Zeq 1.8 + j 1.6

3.2 AC Circuit Analysis

121

Circuit in Fig. 3.27 A circuit with a voltage-controlled current source is shown in Fig. 3.27. Z1 = 2,

Z2 = 2 − j 1,

Z3 = 3.

By nodal analysis, we get −1 − 2V2 + Solving, we get V2 = I1 = −1A,

Z2 1−2Z2

I2 =

V2 = 0. Z2

= −0.6154 − j 0.0769. The currents are

V2 = −0.2308 − j 0.1538A, Z2

I3 = −(I1 + I3 ) = 1.2308 + j 0.1538A

With I1 + I2 + I3 = 0, KCL is satisfied. Mesh Analysis Z2 (2V2 + 1) = V2 ,

or

V2 =

Z2 V 1 − 2Z2

as before. Let us find the current I2 through Z2 using The´venin’s theorem. −1 − 2V2 = 0 and V2 = Voc =

−1 . 2

Let us apply a 1 A current source instead of the load impedance and open-circuit the independent current source. Then, 1 + 2V2 = 0 or Zeq = V2 = −0.5. Therefore, the current through Z2 is I2 =

Voc −0.5 = −0.2308 − j 0.1538 = Z2 + Zeq 1.5 − j 1

as found earlier. Fig. 3.27 A circuit with a voltage-controlled current source

i1 R1

R3 v2 i3 i2 3Ω 2v2 R2 2Ω 2Ω C2 1F

cos(t)A ∼

122

3 AC Circuits

Fig. 3.28 A circuit with a current-controlled current source

i1 R1

v2 i2

R3 3Ω

i3 2i2

2Ω C2

cos(t)A ∼ 1F

R2 2Ω

Circuit in Fig. 3.28 A circuit with a current-controlled current source is shown in Fig. 3.28. Z1 = 2,

Z2 = 0.4 − j 0.8,

Z3 = 3

Nodal Analysis −1 − 2

V2 V2 + = 0. Z2 Z2

Solving, we get V2 = −Z2 = −0.4 + j 0.8. The currents are I1 = −1 A,

I2 = −1 A,

I3 = 2 A.

With I1 + I2 + I3 = 0, KCL is satisfied. Mesh analysis Z2 (2

V2 + 1) = V2 , Z2

or

V2 = −Z2 V

as before.

3.3

Circuit Theorems

Except for the same two differences between AC and DC circuit analysis, the theory remains the same as for DC circuits. The values of circuit elements become complex quantities involving complex arithmetic. The AC circuit must be analyzed separately at each frequency of interest.

3.3.1

Thévenin’s Theorem

Any linear combination of voltage and current sources, independent or dependent, and impedances with two terminals can be replaced by a fixed voltage source Veq or Voc , called Thévenin equivalent voltage, in series with an impedance Zeq , called Thévenin equivalent impedance. The equivalent source is derived as follows:

3.3 Circuit Theorems

123

Fig. 3.29 Thévenin and Norton equivalent circuits

a

Zeq + Voc ∼

cos(t+ π6 )V

+ ∼

C1

3Ω R2

L2 1H

Z1 π 6V

∼

0.1F

Z2

Va I =? Z5 1 Z3 j1

Vb

−j10 1−j10

C4

(a)

b

j3 3+j

−j10

1Ω

ZL (b)

1 R4 1Ω

Zeq

b

R5

L3 1H

IL Isc ∼

ZL (a)

0.1F

a IL

Z4

(b)

Fig. 3.30 A bridge circuit and its frequency-domain version

1. Remove the load circuit, any combination of independent voltage and current sources and impedances and find the open-circuit voltage, called Veq , across the load circuit terminals. 2. Find the impedance across the load terminals, after short-circuiting all independent voltage sources and open-circuiting all independent current sources, called Zeq . This step can also be carried out by finding the short-circuit current Isc through the load circuit terminals and Zeq = Veq /Isc . The´venin equivalent circuit is shown in Fig. 3.29a. The bridge circuit, we analyzed earlier, is shown in Fig. 3.30a. The frequency-domain version of the circuit is shown in Fig. 3.30b. The problem is to find the current I through Z5 using Thévenin’s theorem. The impedance values are Z1 = −j 10,

Z2 =

j3 , 3 + j1

Z3 = j 1,

Z4 =

−j 10 , 1 − j 10

Z5 = ZL = 1.

With the impedance Z5 disconnected in Fig. 3.30b, the voltages Va , Vb and Vab , the open-circuit voltage, are found as Va = V

Z3 = −0.0962 − j 0.0556, Z1 + Z3

Vb = V

Z4 = 0.6496 − j 0.0860, Z2 + Z4

Vab = Va − Vb = −0.7458 + j 0.0305. The equivalent impedance at terminals a and b in Fig. 3.31a, which is the series combination of parallel impedances Z1 ||Z3 and Z2 ||Z4 is determined as Zeq =

Z1 Z3 Z2 Z4 + = 0.5152 + j 1.4589. Z1 + Z3 Z2 + Z4

124

3 AC Circuits

−j10

j3 3+j

Z1

Z2

a Z3 j1

−0.7458+j0.0305V

Fig. 3.31 (a) Circuit to determine the equivalent impedance from ab; (b) circuit showing the determination of the load current

b −j10 1−j10

Z4

0.5152+j1.4589 Zeq ∼

Z5 1

IL = −0.2454+j0.2564

(a)

(b) V1

cos(t+ π6 )V

0.1F + ∼

C1

3Ω R2

L2 1H

R5

Z1 1

1Ω L3 1H

R4 1Ω

j3 3+j

−j10 π 6V

∼

0.1F

Z2

V I ∼ Z3 j1

C4

−j10 1−j10

Z4

V2 (a)

(b)

Fig. 3.32 (a) A bridge circuit; (b) with a bucking voltage V replacing the load impedance

Now, the load current IL is determined as shown in Fig. 3.31b IL =

Vab −0.7458 + j 0.0305 = −0.2454 + j 0.2564, = Zeq + ZL (0.5152 + j 1.4589) + 1

which is the same as found in the complete analysis of the circuit in an earlier example. Bucking Voltage Method The´venin equivalent circuit is characterized by the equation V = I Zeq + Voc . If I = 0, V = Voc , the open-circuit voltage. If V = 0, Isc = − ZVoc , the short-circuit current. Applying eq KVL and KCL at the top and bottom nodes at the right side of Fig. 3.32b, the equilibrium equations are V1 − V2 = (0.8660 + j 0.5) V1 V1 − V V 2 V2 − V + + + = 0. Z2 Z1 Z4 Z3 The second equation is obtained using the fact that the current leaving the bottom node and entering the top node must be equal. Simplifying, we get V1 − V2 = (0.8660 + j 0.5) V1 (Z1 Z3 Z4 + Z2 Z3 Z4 ) + V2 (Z1 Z2 Z3 + Z1 Z2 Z4 ) = V (Z1 Z2 Z4 + Z2 Z3 Z4 ).

3.3 Circuit Theorems

125

cos(t+ π6 )A

0.1F C1

3Ω R2

L2 1H

R5

∼

Z1 1

1Ω L3 1H

R4 1Ω

j3 3+j

−j10 π 6A

∼

0.1F

Z2

Va Z I =? 5 1 Z3 j1

−j10 1−j10

C4

(a)

Vb Z4

(b)

Fig. 3.33 (a) A bridge circuit with a current source; (b) its frequency-domain version

Solving for V1 and V2 in terms of V , we get V1 = V (0.3229−j 0.2391)+(0.4499+j 0.3978) and V2 = V (0.3229−j 0.2391)−(−0.4161−j 0.1022)

The equilibrium equation at node with voltage V is V − V2 V − V1 + = I. Z1 Z3 Substituting for V1 and V2 in this equation, we get V = (0.5152 + j 1.4589)I + (−0.7458 + j 0.0305) = I Zeq + Voc . The first method seems to be easier for this problem. Consider the bridge circuit with a current source shown in Fig. 3.33a and its frequency-domain version shown in Fig. 3.33b. π π I = cos( ) + j sin( ) = (0.8660 + j 0.5) A 6 6 The problem is to find the current through Z5 . With Z5 disconnected, the voltages Va , Vb Vab = Voc are computed as I (Z2 + Z4 ) = −0.1104 − j 0.1459 Va = Z3 Z1 + Z2 + Z3 + Z4 Vb = Z4

I (Z1 + Z3 ) = 1.0405 + j 0.2856 Z1 + Z2 + Z3 + Z4

Vab = Va − Vb = (−1.1509 − j 0.4315)V In order to find Zeq , the current source is replaced by an open-circuit. Then, we get Zeq =

(Z1 + Z2 )(Z3 + Z4 ) = (1.1993 + j 0.8475) . Z1 + Z3 + Z2 + Z4

126

3 AC Circuits

2 cos(5t)A V3 ∼ I I3 2 2Ω

I1 Z1 jω0.1

Z2

sin(t+π3 )V

V1 1 − π6 + ∼ Z

3

1 1+jω0.2

10

+ 2V

Z6 = 1+jω0.2

Z4

V2

2 Z5 = 1+j2ω0.1

Fig. 3.34 A circuit with voltage and current sources with different frequencies

Therefore, the current through Z5 is IL =

Vab −1.1509 − j 0.4315 = −0.5215 + j 0.0048. = Zeq + ZL (1.1993 + j 0.8475) + 1

Consider the circuit with voltage and current sources with different frequencies, shown in Fig. 3.34. The problem is to find the current through the impedance Z5 . Let us find the current due to the current source I = 2 cos(5t) = 2 0 A alone, with the voltage sources short-circuited. With ω = 5, Z5 = 1 − j 1,

Z6 = 1 + j 1.

The voltages at the top and bottom terminals of Z5 and across, with Z5 disconnected, are Va = −2(1 + j 1),

Vb = 0 and

Vab = Va − Vb = −2(1 + j 1).

In order to find Zeq , we replace the two voltage sources by short-circuits and the current source by open-circuit. Then, Zeq = Z6 = 1 + j 1. Due to the AC current source alone, the current through Z5 is IL =

Vab −2(1 + j 1) = −1 − j 1. = Zeq + ZL (1 − j 1) + (1 + j 1)

Due to the rightmost DC source alone 2 A. 3 The current source is open-circuited and the other voltage source is short-circuited. The sum of the two currents is the total current in RL , IL =

2 √ 3π + 2 cos(5t − ). 3 4 No contribution from the leftmost voltage source, since the current source becomes open-circuited.

3.3 Circuit Theorems

127

2 cos(5t)A V ∼

2Ω

sin(t+π3 )V

Z1 jω0.1

Z2

1 1+jω0.2

+ ∼ Z

3

10

I ∼

+ 2V

Z6 = 1+jω0.2

Z4

2 Z5 = 1+j2ω0.1

Fig. 3.35 A circuit with voltage and current sources with different frequencies with a bucking voltage inserted

Bucking Voltage Method Circuit analysis using Thévenin’s theorem is a special case of nodal analysis, characterized by the equation V = I Zeq + Veq . Impedance Z5 has been replaced by a current source with current I and voltage V at its top terminal, as shown in Fig. 3.35. Applying KCL at node with voltage V for the current source alone, we get I−

V =2 1 + j1

(3.44)

V = I (1 + j 1) − 2(1 + j 1) Vab = Veq = −2(1 + j 1),

Zeq = 1 + j 1

Due to the AC current source alone IL =

Vab −2(1 + j 1) = (−1 − j 1) A = Zeq + ZL (1 − j 1) + (1 + j 1)

as found before. Due to the DC source alone I =

V −2 1

or

Vab = Veq = 2, and

V =I +2 Zeq = 1

2 2 = A 2+1 3 Consider the circuit with a voltage-controlled voltage source, shown in Fig. 3.36a. The problem is to find the current through Z2 = 1 . IL =

Nodal Analysis With Z1 = j 1,

Z2 = 1,

Z3 = 5 − j 10,

V = cos(t)

(V2 − 1) V2 V2 + 2(V2 − 1) + + =0 Z1 Z2 Z3

128

3 AC Circuits

v2

L 1H

0.1F

C

R3 5Ω

v1 +R2 1Ω ∼ cos(t)V 2(v2 −v1 )

L 1H v1 ∼ cos(t)V

+ −

(a)

0.1F R3 5Ω

C L 1H

0.1F

+v ∼ cos(t)V

+ −

C

v2

2(v2 −v1 ) (b)

R3 5Ω

+ −

v2

2(v2 −v1 ) (c)

Fig. 3.36 (a) A circuit with a voltage-controlled voltage source; (b) with the load resistor disconnected; (c) the load resistor replaced by a voltage cos(t)

V2 =

Z2 Z3 + 2Z1 Z2 = 0.3974 − j 0.4803 Z2 Z3 + Z1 Z3 + 3Z1 Z2 V2 = 0.3974 − j 0.4803. Z2

IZ2 =

Thévenin’s Theorem Approach The circuit is shown in Fig. 3.36b with the load resistor disconnected. Let us find the open-circuit voltage at the terminals shown by small discs. Applying KCL at node V2 , we get (V2 − 1) V2 + 2(V2 − 1) + =0 Z1 Z3 V2 = Veq =

Z3 + 2Z1 = 1.0946 − j 0.0676. Z3 + 3Z1

In order to find the short-circuit current, we turn off all independent sources only. Then, a voltage source of 1 (0) V is applied at the open-circuited load terminals, as shown in Fig. 3.36c. The resultant current in that branch is found. Then, the inverse of that current is Zeq . Applying 1 0 V, i=

1 1+2 + = 0.1200 − j 0.7600 Z1 Z3

1 = 0.2027 + j 1.2838. i For linearity and Thévenin’s theorem, leave the dependent sources, as they are controlled by circuit variables. Alternatively, applying a current of 1 0 A, Zeq =

V2 V2 + 2V2 + =1 Z1 Z3 V2 (3Z1 + Z3 ) = Z1 Z3 V2 = Zeq =

Z1 Z3 = 0.2027 + j 1.2838. (3Z1 + Z3 )

3.3 Circuit Theorems

129

v2

v2

v2 0.1F R3 5Ω

v1 R2 1Ω ∼ cos(t)V 2(v2 −v1 )

R2 1Ω

v1 R2 1Ω ∼ cos(t)V 2(v2 −v1 )

+ −

(a)

+V ∼ cos(t)V

L 1H

L 1H + −

L 1H

+ −

C

2(v2 −v1 )

(b)

(c)

Fig. 3.37 (a) A circuit with a voltage-controlled voltage source; (b) with the load impedance disconnected; (c) the load impedance replaced by a voltage source cos(t)

Bucking Voltage Method (V − 1) V + 2(V − 1) + =I Z1 Z3 V (3Z1 + Z3 ) = (Z3 + 2Z1 ) + I Z1 Z3 V =

(Z3 + 2Z1 ) + I Z1 Z3 = (1.0946 − j 0.0676) + I (0.2027 + j 1.2838) (3Z1 + Z3 ) IL =

(1.0946 − j 0.0676) = 0.3974 − j 0.4803. Z2 + (0.2027 + j 1.2838)

Let us find the current through the resistor and capacitor in the circuit, shown in Fig. 3.37a, which is the same as the last one. The circuit is shown in Fig. 3.37b with the load impedance disconnected. By nodal analysis, in the last example, we found V2 = Va = 0.3974 − j 0.4803 Vb = −2(V2 − 1) = 1.2052 + j 0.9607 IL =

Va − Vb = 0.0830 − j 0.1223. Z3

Thévenin’s Theorem Approach With the load impedance disconnected, let us find the open-circuit voltage. Applying KCL at node V2 , we get V2 (V2 − 1) + =0 Z1 Z2 V2 =

Z2 = 0.5 − j 0.5. Z1 + Z2

By voltage division also, we get the result. Voc = V2 − (−2(V2 − 1)) = 3V2 − 2 = (−0.5 − j 1.5). Let us find Zeq . The circuit is shown in Fig. 3.37c. Applying 1 0 V, V2 − 1 = −2V2

or

V2 =

1 . 3

130

3 AC Circuits

The current through the 1 − V source, which is the sum of the currents through Z1 and Z2 , is 1 + j1 1 1 ( + 1) = . 3 j1 j3 The inverse of it is Zeq , Zeq =

j3 = j 1.5(1 − j 1) = 1.5 + j 1.5. 1 + j1

Alternatively, applying 1 0 A, V2 =

j1 . 1 + j1

The voltage across the 1 − A source is V2 + 2V2 = 3V2 = Zeq =

j3 = j 1.5(1 − j 1) = 1.5 + j 1.5. 1 + j1

With Z3 = 5 − j 10, Zeq = 1.5 + j 1.5 and Voc = −0.5 − j 1.5, IL =

3.3.2

−0.5 − j 1.5 = 0.0830 − j 0.1223 A. Z3 + (1.5 + j 1.5)

Norton’s Theorem

Admittance is the reciprocal of the impedance. E = ZI,

I = YE

The first form is more convenient to find E, when I (the excitation) is given. The second form is more convenient to find I , when E (the excitation) is given. Any linear combination of voltage and current sources, independent or dependent, and impedances with two terminals can be replaced by a fixed current source Isc , called Norton equivalent current, in parallel with an impedance Zeq . The equivalent source is derived as follows: 1. Remove the load circuit, any combination of independent voltage and current sources and impedances. and find the short-circuit current, called Isc , through the load circuit terminals. 2. Find the impedance across the load terminals, after short-circuiting all independent voltage sources and open-circuiting all independent current sources, called Zeq . This step can also be carried out by finding the short-circuit current Isc through the load circuit terminals and the open-circuit voltage Voc . Then, Zeq = Voc /Isc . Figure 3.29b shows the Norton equivalent circuit. Norton equivalent circuit is characterized by the equation V − Isc = V Yeq − Isc . I= Zeq It is better use Thévenin’s theorem for small Zeq and Norton’s theorem for large Zeq compared with other impedances in the circuit. Once we find Voc and Zeq , as showed for Thévenin’s equivalent circuit, then Isc = Voc /Zeq .

3.3 Circuit Theorems

3.3.3

131

Maximum Average Power Transfer Theorem

An AC circuit to find the condition for maximum average power transfer between a source and a load is shown in Fig. 3.38. A load impedance ZL = RL +j XL is connected to the AC source and its source impedance is Zs = Rs +j Xs . The value of ZL , for which the maximum average power transfer occurs from the source to the load, is to be found. The current through the load impedance is I=

V . Zs + ZL

The average power absorbed by the load, as shown in a later chapter, is PL = 0.5|I |2 RL = 0.5

|V |2 RL . (Rs + RL )2 + (Xs + XL )2

Intuitively, as (Xs + XL ) absorbs no power, it must be zero for maximum power transfer. That is, XL = −Xs . Differentiating this expression with respect to XL and equating to zero, we get dPL = |V |2 RL dXL



−(Xs + XL ) ((Rs + RL )2 + (Xs + XL )2 )2

 =0

For PL to be maximum, XL = −Xs . Substituting this condition in the expression for power, we get PL = 0.5

|V |2 RL . (Rs + RL )2

This expression is the same as that obtained for DC circuits and yields the maximum power with RL = Rs . The maximum average power delivered to the load is PL =

|V |2 . 8RL

Therefore, the condition for maximum average power transfer is that the source and load impedances are conjugate symmetric, ZL = Zs∗ . When this condition is satisfied, the total impedance of the circuit becomesresistive. For a purely resistive load RL , the condition for maximum power transfer is RL = |Zs | = Rs2 + Xs2 . Fig. 3.38 An AC circuit to find the condition for maximum average power transfer

I V

∼

Zs Rs+jXs RL+jXL ZL

132

3 AC Circuits

Example Find the load impedance Z2 in the bridge circuit, shown in Fig. 3.39a, for maximum power transfer from the source to the load. The source voltage is 0.8660 + j 0.5. We have to find the Thévenin’s equivalent voltage and impedance across the load impedance terminals. Replacing the bottom half of the Δ circuit by Y circuit, we get, with Zbc = Z4 = 0.9901 − j 0.0990,

Zac = Z3 = j 1,

Zab = Z5 = 1.

Za =

j1 × 1 = 0.1888 + j 0.4170 (0.9901 − j 0.0990) + j 1 + 1

Zb =

(0.9901 − j 0.0990) × 1 = 0.3942 − j 0.2282 (0.9901 − j 0.0990) + j 1 + 1

Zc =

(0.9901 − j 0.0990) × j 1 = 0.2282 + j 0.3942. (0.9901 − j 0.0990) + j 1 + 1

The transformed circuit, with the voltage source short-circuited, is shown in Fig. 3.39b. Now, Zeq = (0.3942−j 0.2282)+

(−j 10 + (0.1888 + j 0.4170))(0.2282 + j 0.3942) = 0.6425+j 0.1763 ((−j 10 + (0.1888 + j 0.4170)) + (0.2282 + j 0.3942))

∗ = 0.6425 − j 0.1763. The V across The maximum average power transfer occurs with ZL = Zeq oc the load terminals of the transformed circuit, with the voltage source inserted, is

Voc =

((−j 10 + (0.1888 + j 0.4170)))(0.866 + j 0.5) = 0.9155+j 0.4977 = 1.0420 (0.4979) −j 10 + (0.1888 + j 0.4170) + (0.2282 + j 0.3942)

Therefore, the maximum power transferred is Pm =

cos(t + π6 )V

0.1F + ∼

C1

|Voc |2 (1.04202 ) = = 0.2113 W. 8RL (8 × 0.6425)

3Ω R2

L2 1H

R5 1Ω

L3 1H

R4 1Ω

0.1F C4

(a) Fig. 3.39 (a) A bridge circuit; (b) circuit to find Zeq

−j10 Zeq = ZL∗ b a 0.3942−j0.2282 0.1888+j0.4170 0.2282+j0.3942 c (b)

3.3 Circuit Theorems

133

P

0.21125

0.6

0.6425

0.8

RL

Fig. 3.40 Load resistor versus power transferred

cos(t+ π6 )A ∼ v2 i R1

C

0.2F

1Ω

cos(t+ π6 )A ∼

i1 R1

C

0.2F

1Ω

R1

L 0.1H

L 0.1H

∼ 2 sin(t+ π4 )A (a)

∼ 2 sin(t+ π4 )A (b)

v1

i2

C

0.2F

−j5 I

+∼ 5 − π3 V

1

1Ω L 0.1H

j0.1 0.2 π4 V ∼ +

(c)

(d)

Fig. 3.41 (a) A circuit with current sources; (b) and (c) the circuits with one of the two current sources replaced by an open-circuit; (d) equivalent circuit with voltage sources in the frequency-domain

Figure 3.40 show the power transferred for a range of values of the load resistor. The peak value occurs when the load resistance is equal to the Thévenin equivalent resistance.

3.3.4

Source Transformation

Thévenin and Norton equivalent circuits provide the same volt–ampere relationship at the terminals. It is possible to substitute one source by another, called source transformation, to simplify the circuit analysis. Source transformation is most useful if the source is localized to some portion of the circuit. Consider the circuit shown in Fig. 3.41a with current sources. The problem is to find the current through R1 . The sources are 2 cos(t − π4 )A and cos(t + π6 )A. In the frequency-domain, they are 2 (− π4 ) and 1 ( π6 ). Nodal Method Applying KCL at the right side nodes, we get, with ZL = j 0.1, ZC = −j 5 and ZR1 = 1,  π V1 − V2 V1 + = −2 − j 0.1 1 4 π  V2 − V1 V2 + = 1 . −j 5 1 6

134

3 AC Circuits

Simplifying, we get

π  V1 (1 + j 0.1) − j 0.1V2 = −0.2 4  π V1 (j 5) + (1 − j 5)V2 = 5 − 3      V1 1 + j 0.1 −j 0.1 −0.1414 − j 0.1414 . = j5 1 − j5 V2 2.5000 − j 4.3301

The determinant of the admittance matrix is 1 − j 4.9. Solving the equilibrium equations, we get 

V1 V2



 =

−1 

j5 1 − j5 

=

−0.1414 − j 0.1414

−0.1764 − j 0.0488 0.7498 + j 0.3012



2.5000 − j 4.3301

1.0196 − j 0.0040 −0.0196 + j 0.0040 0.9796 − j 0.1999

 =

1 + j 0.1 −j 0.1



−0.1414 − j 0.1414



2.5000 − j 4.3301

0.0204 + j 0.1999  .

Since the resistance value is 1 , the current through it, from bottom to top, is V1 − V2 = (−0.9263 − j 0.35) A. Let us find the current using the linearity property. The currents due to each of the two current sources, I1 in Fig. 3.41b and I2 in Fig. 3.41c, are found separately and added to find the total current. When one of the sources is open-circuited, the circuit becomes a current source feeding a parallel circuit. In Fig. 3.41b, the equivalent impedance is (j 0.1)  (1 − j 5). In Fig. 3.33c, the equivalent impedance is (−j 5)  (1 + j 0.1). The currents can be found by current division. I1 = −

(j 0.1)(2(cos(− π4 ) + j sin(− π4 ))) = 0.0221 − j 0.0334 (1 + j 1 − j 5)

I2 = −

(−j 5)(cos( π6 ) + j sin( π6 )) = −0.9483 − j 0.3167. (1 + j 0.1 − j 5)

Now, I = I1 + I2 = (−0.9263 − j 0.35) A, as found by nodal analysis. Consider the transformed circuit shown in Fig. 3.41d with voltage sources. It is a series circuit. The current sources are multiplied by the respective shunt impedances to get the equivalent voltage sources and the shunt impedances becomes the series impedances for the voltage sources. In the frequency-domain, the voltage sources are 2(0.1)

 π π = 0.1414 + j 0.1414 − + 4 2

and

1(5)

π 6

−

π = 2.5 − j 4.3301. 2

The current through the resistance is the sum of the voltages divided by the three impedances in series 2.6414 − j 4.1887 = (0.9263 + j 0.35) A. 1 + j 0.1 − j 5

3.4 Application

135

This current flows in the opposite direction. The current is I = (−0.9263 − j 0.35) A, as obtained earlier. With voltage sources, we got the result by using the simpler Ohm’s law.

3.4

Application

3.4.1

Filters

A practical signal is composed of different frequency components. At the time of generation or in transmission, desired signals get contaminated with undesirable signals, called noise signals. Typically, the desired signals are composed of low frequency components and noise signals are composed of high frequency components. Filters are commonly used in signal processing applications to filter out the unwanted noise component. Basic lowpass filters, which pass low frequency components and suppress high frequency components, are presented.

RL Filter The RL lowpass filter is shown in Fig. 3.42a. The input signal to the filter is v1 (t) = cos(10t) + 0.1 cos(1000t). It is desired that the output v2 (t) contains the high frequency component of magnitude less than 10% of that at the input. By voltage division, the input–output relationship of the filter, in the frequency domain, is V2 R . = V1 R + j ωL The output, for a given input is inversely proportional with the frequency due to the frequency ω appearing in the denominator. The magnitude has to meet the constraint     R    R + j ωL  = 0.1. Substituting for the magnitude, we get R  = 0.1 2 R + (ωL)2

or

100R 2 = R 2 + (ωL)2 .

With ω = 1000 and R = 1000 , solving for L, we get L = 9.9499 H. The magnitude of the transfer function, the input–output relationship in the frequency-domain, is shown in Fig. 3.43a. It

+

+

L

v1 (t)

R

v2 (t)

+ v1 (t)

(a) Fig. 3.42 (a) A RL lowpass filter circuit; (b) a RC lowpass filter circuit

+

R C

(b)

v2 (t)

136

3 AC Circuits

1

x(t)

|H(j )|

1 0.9951

0.1 10

0

-1 0

1000

0.6283

t

(a)

(b)

Fig. 3.43 (a) Magnitude of the transfer function of the filter; (b) excitation and the response of the RL circuit

is a plot of the magnitude and the phase (not shown) of the filter output for the input ej ωt with the frequency varying. Since the filter is a lowpass one, it attenuates the high frequency components more. At ω = 10 rad/s, the attenuation is 0.9951. At ω = 1000 rad/s, the attenuation is 0.1. While the attenuation of the high frequency component is desirable, the low frequency components are also attenuated to a lesser extent. Filters are designed to suit the application requirements. The input and the output waveforms of the filter are shown in Fig. 3.43b. The period of the low frequency component is 2π/10 = 0.6283 s. The output waveform, shown by the dashed line, has negligible undesirable high frequency component of the input. The filter circuit is a variable frequency voltage divider, since the impedance of the inductor increases with increasing frequency. Therefore, most of the voltage drop due to high frequency components occur across it and, thereby, the filter output, the voltage across the resistor, contains a low percent of the high frequency components. The output waveform is shifted due to the filtering operation.

RC Filter The RC lowpass filter is shown in Fig. 3.42b. The input signal to the filter is V1 (t) = cos(10t) + 0.1 cos(1000t). It is desired that the output v2 (t) contains the high frequency component of magnitude less than 10% of that at the input. By voltage division, the input–output relationship of the filter, in the frequency domain, is V2 1 . = V1 1 + j ωRC The magnitude has to meet the constraint     1    1 + j ωRC  = 0.1. Substituting for the magnitude, we get 1 = 0.1 1 + (ωRC)2

or

100 = 1 + (ωRC)2 .

With ω = 1000 and R = 1000 , solving for C, we get C = 9.9499 µF. The transfer function and the output are similar to that in Fig. 3.43.

3.5 Summary

3.5

137

Summary

• Alternating current (AC) is also movement of electrical charge, but changes direction periodically. • A sinusoidal waveform is a linear combination of sine and cosine waveforms. • System models, such as differential equation and convolution, reduce to algebraic equations for a sinusoidal input for linear systems. • The polar form of a sinusoid is x(t) = A cos(ωt + θ ), −∞ < t < ∞ • The rectangular form of a sinusoid is A cos(ωt + θ ) = C cos(ωt) + D sin(ωt). −∞ < t < ∞ is found to be • The complex sinusoid, v(t) = V ej (ωt+θ) = V ej θ ej ωt , indispensable for analysis due to its compact form and ease of manipulation of the exponential function. Due to Euler’s identity, v(t) =

 V  j (ωt+θ) e + e−j (ωt+θ) = V cos(ωt + θ ) 2

• AC circuit analysis boils down to analysis with sources of the form Aej (ωt+θ) , which is a mathematical equivalent of the real sinusoid. Then, the output to the real sinusoid is deduced from the results. • When the complex exponential source is used, the circuit elements, resistors, inductors, and capacitors, are to be represented appropriately. • When the independent variable of signals, such as time t, is not frequency, then that representation is called the time-domain representation. On the other hand, when the independent variable of signals is frequency, such as f and ω, then that representation is called the frequency-domain representation. • The first task in formulating the equilibrium equations for an electric circuit is to select a set of variables, voltages or currents. They must be independent and adequate to describe the state of the network at any instant. They must be uniquely and reversibly related to all the branch variables. • After selecting an appropriate set of variables, we have to use Kirchhoff’s laws. • AC circuits must be analyzed at each frequency individually, if sources with more than one frequency are present in the circuit. • As the sinusoidal waveform is everlasting, using a sinusoid as a source will result in the steady-state analysis. • Any linear combination of voltage and current sources, independent or dependent, and impedances with two terminals can be replaced by a fixed voltage source Veq , called Thévenin’s equivalent voltage, in series with an impedance Zeq , called Thévenin’s equivalent impedance. • Any linear combination of voltage and current sources, independent or dependent, and impedances with two terminals can be replaced by a fixed current source Isc , called Nortan’s equivalent current, in parallel with an impedance Zeq . • The condition for maximum power transfer is that the source and load impedances are conjugate symmetric, ZL = Zs∗ When this condition is satisfied, the total impedance of the circuit becomes resistive. • Source transformation is replacing a current source by a voltage source and vice versa for simpler circuit analysis. It is most useful if the source is localized to some portion of the circuit.

138

3 AC Circuits

Exercises 3.1 Find the sinusoid in cosine form. Find the value of t closest to t = 0, where the first positive peak of the sinusoid occur. π 3.1.1 x(t) = sin( 2π 8 t + 4 ). 2π 3.1.2 x(t) = 2 sin( 6 t − π6 ). * 3.1.3 x(t) = −5 sin(2π t + π3 ). 3.2 Find the rectangular form of the sinusoid. Find the value of t closest to t = 0, where the first positive peak of the sinusoid occur. Find the values of t at which the next two consecutive positive peaks occur. Get back the polar form from the rectangular form and verify that the given sinusoid is obtained. π 3.2.1 x(t) = sin( 2π 6 t + 6 ). 2π 3.2.2 x(t) = 3 cos( 3 t − π3 ). 3.2.3 x(t) = −2 sin(2π t + 2π 3 ). * 3.2.4 x(t) = 4 sin(2π t − π4 ). 2π 3.2.5 x(t) = 5 cos( 2π 4 t + 6 ). 3.3 Express the signal in terms of complex exponentials. 3.3.1 x(t) = sin( 2π 8 t) 3.3.2 x(t) = cos(2π t) π 3.3.3 x(t) = 2 cos( 2π 4 t − 3) 2π π *3.3.4 x(t) = 3 sin( 6 t + 6 ) 3.4 Find the current i(t) in the series circuit in cosine sinusoidal form, using the frequency-domain representation. 3.4.1 π 2π v(t) = 2 cos( t − ), R = 3, C = 0.2 8 3 * 3.4.2 π 2π v(t) = 2 sin( t − ), R = 2, L = 0.1 4 3 3.5 Find the current I and the voltages across the impedances, VZ1 , VZ2 , VZ3 , VZ4 , VZ5 in the series circuit, shown in Fig. 3.44. Verify KVL. 3.5.1 v(t) = 3 cos(2t)V , Z3 =

(2)(j ω0.1) , (2 + j ω0.1)

Z1 =

1 , j ω0.3

Z4 = 1,

Fig. 3.44 Series circuit with a voltage source

Z2 = 2,

Z5 = j ω0.2

VZ1 + V ∼

Z1

VZ2 I

Z2 Z3

VZ5

VZ4

Z5

Z4

VZ3

Exercises

139

Fig. 3.45 Series circuit with a current source

VZ1

VZ2 I

Z1

I ∼

Z2 Z3

VZ5

VZ4

Z5

Z4

* 3.5.2 v(t) = 2 sin(3t)V , Z3 =

Z1 =

(2)(j ω0.2) , (2 + j ω0.2)

1 , j ω0.1

Z4 = 2,

Z2 = 3,

Z5 = j ω0.3

3.5.3 v(t) = cos(2t +

π )V , 3

(3)(j ω0.1) , (3 + j ω0.1)

Z3 =

Z1 =

1 , j ω0.2

Z4 = 1,

Z2 = 1,

Z5 = j ω0.4

3.6 Find the voltages across the impedances, VZ1 , VZ2 , VZ3 , VZ4 , VZ5 in the series circuit, shown in Fig. 3.45. 3.6.1 i(t) = sin(2t +

π )A, 3

1 ) (4)( j ω0.1

Z3 =

1 (4 + ( j ω0.1 ))

Z1 =

1 , j ω0.1

Z4 = 3,

,

Z2 = 2,

Z5 = j ω0.1

* 3.6.2 i(t) = 2 sin(2t)A, Z3 =

Z1 =

1 ) (3)( j ω0.1 1 (3 + ( j ω0.1 ))

,

1 , j ω0.2

Z4 = 4,

Z2 = 1, Z5 = j ω0.1

3.6.3 i(t) = cos(2t − Z3 =

π )A, 3

1 ) (3)( j ω0.1 1 (3 + ( j ω0.1 ))

,

Z1 =

1 , j ω0.1

Z4 = 3,

Z2 = 2,

Z5 = j ω0.4

VZ3

140

3 AC Circuits

Fig. 3.46 Parallel circuit with a current source

IZ1 IZ2 IA ∼

Fig. 3.47 Parallel circuit with a voltage source

I + V ∼

Z1

Z2

IZ1 IZ2 Z1

Z2

IZ3 Z3

IZ3 Z3

IZ4 IZ5 Z4

IZ4 IZ5 Z4

3.7 Find the voltage across the impedances, VZ1 , VZ2 , VZ3 , VZ4 , VZ5 in the parallel circuit, shown in Fig. 3.46. Find the currents through the impedances, IZ1 , IZ2 , IZ3 , IZ4 , IZ5 Verify that the sum of the currents through the impedances is equal to the source current. 3.7.1 π )A, Z1 = 1, Z2 = j ω0.1, 4 1 , Z5 = 3 Z4 = j ω0.2

i(t) = sin(2t + Z3 = 2, 3.7.2 i(t) = 2 cos(t)A, Z4 =

1 , j ω0.1

Z1 = 3,

Z2 = j ω0.2,

Z3 = 2,

Z5 = 1

* 3.7.3 π )A, Z1 = 1, Z2 = j ω0.1, 6 1 , Z5 = 3 Z4 = j ω0.2

i(t) = cos(t + Z3 = 2,

Z5

3.8 Find the current I and the voltages across the impedances, VZ1 , VZ2 , VZ3 , VZ4 , VZ5 in the parallel circuit, shown in Fig. 3.47. Find the currents through the impedances, IZ1 , IZ2 , IZ3 , IZ4 , IZ5 Verify that the sum of the currents through the impedances is equal to the total current, I .

Z5

Exercises

141

3.8.1 π )V , 6

v(t) = 2 cos(t + Z4 =

1 , j ω0.3

Z1 = 2,

Z2 = j ω0.2,

Z3 = 1,

Z1 = 3,

Z2 = j ω0.3,

Z3 = 2,

Z1 = 2,

Z2 = j ω0.2,

Z3 = 1,

Z5 = 4

* 3.8.2 π )V , 4

v(t) = 2 sin(2t + Z4 =

1 , j ω0.2

Z5 = 3

3.8.3 v(t) = 3 cos(3t − Z4 =

1 , j ω0.3

π )V , 3

Z5 = 4

3.9 Find the current I and the voltages across the impedances, VZ1 , VZ2 , VZ3 , VZ4 , VZ5 and current through IZ1 , IZ2 , IZ3 , IZ4 , IZ5 in the series-parallel circuit, shown in Fig. 3.48. Verify KVL around the loops and KCL at node x. 3.9.1 v(t) = cos(2t)V , Z3 = j ω0.3,

1 , j ω0.3

Z1 =

Z4 = 2,

Z2 = 3,

Z5 = 1

* 3.9.2

Fig. 3.48 Series-parallel circuit with a voltage source

1 , j ω0.1

v(t) = sin(t)V ,

Z1 =

Z3 = j ω0.2,

Z4 = 3,

Z2 = 2,

Z5 = 2

Z5 + V ∼

IZ1 Z1

IZ2 IZ3 Z2

Z3

xI IZ4 Z4

142

3 AC Circuits

Fig. 3.49 Series-parallel circuit with a current source

IZ1

I ∼

Z5 xI IZ2 IZ3 IZ4

Z1

Z2

Z3

Z4

3.9.3 π )V , 6

v(t) = cos(2t + Z3 = j ω0.1,

Z1 =

Z4 = 3,

1 , j ω0.2

Z2 = 4,

Z5 = 3

3.10 Find the voltages across the impedances, VZ1 , VZ2 , VZ3 , VZ4 , VZ5 in the series-parallel circuit, shown in Fig. 3.49. Verify KCL at node x. 3.10.1 i(t) = 2 cos(2t + Z3 = j ω0.2,

π )A, 3

Z4 = 1,

Z1 =

1 , j ω0.3

Z2 = 2,

Z5 = 3

* 3.10.2 i(t) = cos(t − Z4 = 4,

π )A, 6

Z1 =

1 , j ω0.2

Z2 = 3,

Z3 = j ω0.3,

Z5 = 1

3.10.3 i(t) = 3 sin(2t + Z4 = 1,

π )A, 6

Z1 =

1 , j ω0.4

Z2 = 1,

Z3 = j ω0.2,

Z5 = 4

3.11 Given a circuit diagram with the independent currents, find the corresponding tree and the loops. Analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. 3.11.1 The circuit is shown in Fig. 3.50. * 3.11.2 The circuit is shown in Fig. 3.51. 3.11.3 The circuit is shown in Fig. 3.52. 3.12 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. 3.12.1 The circuit is shown in Fig. 3.53. * 3.12.2 The circuit is shown in Fig. 3.54.

Exercises

143

i1

Fig. 3.50 Circuit for Exercise 3.11.1

sin(2t− π3 )V

i2 0.1F C1

+ ∼

3Ω R2

R5 1Ω

L3 1H

R4 1Ω i3

i1

Fig. 3.51 Circuit for Exercise 3.11.2

C1

+ ∼

0.1F C4

i2 0.1F

sin(t)V

L2 1H

3Ω R2

L2 1H

R5 2Ω

L3 1H

i3 R4 1Ω

0.1F C4

i1

Fig. 3.52 Circuit for Exercise 3.11.3

cos(t)V

i2 0.1F C1

+ ∼

3Ω R2

L2 1H

R4 1Ω

C4 0.1F

R5 3Ω

L3 1H i3

Fig. 3.53 Circuit for Exercise 3.12.1

i1 1Ω R6

i2 1H L1

sin(2t− π3 )V

+∼ 1H L6

C3

0.1F

2Ω R2

R5 2Ω R4 3Ω i3

L2 1H

0.1F C4

144 Fig. 3.54 Circuit for Exercise 3.12.2

3 AC Circuits

i1 1Ω R6

1H L1

C3

1H L6

Fig. 3.55 Circuit for Exercise 3.13.1

cos(t− π3 )V +∼ R5 3Ω i3 R4 0.1F 3Ω

i1 1Ω R5

R3

sin(2t+ π3 )V +∼ i3 2Ω

1H L5

0.1F C4

2Ω R2

R4 3Ω

0.1F C4

i2

i1 1Ω R5

L2 1H

i2 1H L1

1H L5

Fig. 3.56 Circuit for Exercise 3.13.2

i2 2Ω R2

1H L1

R3

cos(t− π6 )V +∼ 2Ω

3Ω R2

R4 3Ω

0.1F C4

i3 3.13 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. 3.13.1 The circuit is shown in Fig. 3.55. * 3.13.2 The circuit is shown in Fig. 3.56. 3.14 Analyze the circuit using the linearity property. 3.14.1 The circuit is shown in Fig. 3.57. * 3.14.2 The circuit is shown in Fig. 3.58. 3.15 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. 3.15.1 The circuit is shown in Fig. 3.59. * 3.15.2 The circuit is shown in Fig. 3.60. 3.16 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL.

Exercises

145

2 cos(2t)A V3 ∼ I I3 2 2Ω 1 1+jω0.2

sin(t)V

V1 −j + ∼ Z 10 3

1V

cos(t)V

3

2 sin(2t)V

i1

+ ∼

Z2 1 1+jω0.2 10

L1 0.1H

V3

I2

2Ω

Z7 1

I3 2

I1 Z1 jω0.1 V1 1 + ∼ Z

Fig. 3.59 Circuit for Exercise 3.15.1

Z4

V2

sin(2t)A Z5= 1+j2ω0.1

Fig. 3.58 Circuit for Exercise 3.14.2

+ 1V

∼

Z6=jω0.2

Z2

Z7 1

v2 i2

i3 C2 0.2F R2 1Ω R3 2Ω

∼

cos(2t)A

I1 Z1 jω0.1

Z6=jω0.2

Z4

V2

2 Z5= 1+j2ω0.1

Fig. 3.57 Circuit for Exercise 3.14.1

3.16.1 The circuit is shown in Fig. 3.61. * 3.16.2 The circuit is shown in Fig. 3.62. 3.17 Given a circuit diagram with the independent currents, analyze the circuit by both nodal and loop methods to find the voltages and currents at all parts of the circuit. Verify that the results are the same by both the methods. Verify the results using KVL and KCL. 3.17.1 The circuit is shown in Fig. 3.63. * 3.17.2 The circuit is shown in Fig. 3.64. 3.18 Find the Thévenin and Norton equivalent circuits and determine the current through the load impedance. 3.18.1 The load impedance is Z1 in Fig. 3.13a. 3.18.2 The load impedance is Z3 in Fig. 3.14a.

146

3 AC Circuits

L1 0.1H

i2

i3 C2 0.2F R2 1Ω R3 2Ω

∼

v1 i1

+ −

Fig. 3.61 Circuit for Exercise 3.16.1

v2

(v1 −v4 )

R1 2Ω

R5 2Ω R2 1Ω

∼ sin(t)A

Fig. 3.62 Circuit for Exercise 3.16.2

i1 v1

+ −

L1 0.1H

2(v1 −V4 )

R1 2Ω

L1 0.1H

C3

R5 2Ω R2 1Ω

+ ∼

2 cos(t)V v3 + ∼ v4 i2 i3

R3

v2

sin(t)V

i1

v2

2 cos(t)A

Fig. 3.60 Circuit for Exercise 3.15.2

i2

C3 ∼ 2 cos(t)A R3

0.1F

R4

2Ω

3Ω

sin(t)V v3 + ∼ 0.1F

i3

R4

v4 3Ω

3Ω

3.18.3 The load impedance is Z3 in Fig. 3.15a. 3.18.4 The load impedance is Z1 in Fig. 3.16a. 3.18.5 The load impedance is R3 in Fig. 3.17a. 3.18.6 The load impedance is Z1 in Fig. 3.19. 3.38.7 The load impedance is Z3 in Fig. 3.20. 3.19 Find the value of the impedance Z for maximum power transfer from the source to the impedance Z.

Exercises Fig. 3.63 Circuit for Exercise 3.17.1

147

i1

v1

R5 1Ω R2

ix

1Ω

L1 0.1H

2ix

v2 i2

L3 0.1H

R4

v3

4Ω

i3

cos(t)V ∼ +

R3 3Ω

R1 2Ω

Fig. 3.64 Circuit for Exercise 3.17.2

i1

v1 ix

L1 0.1H

R5 1Ω R2

v2

2 Ω i2 2ix

R4 3Ω

v3 i3 + sin(t)V ∼

R3 1Ω

L3 0.1H

R1 3Ω

Fig. 3.65 Circuit for Exercise 3.19.1

2Ω cos(t)V + ∼ Z

3Ω

1F 4Ω

3.19.1 The circuit is shown in Fig. 3.65. * 3.19.2 The circuit is shown in Fig. 3.66. 3.20 Find the current through the impedance Z3 by analyzing the circuit as shown and also after source transformation. * 3.20.1 The circuit is shown in Fig. 3.67. 3.20.2 The circuit is shown in Fig. 3.68.

148

3 AC Circuits

Fig. 3.66 Circuit for Exercise 3.19.2

Z

3Ω

cos(t)V + ∼ 2Ω

Fig. 3.67 A circuit with current sources

1F

∼

4Ω

sin(t)A

Z2 2Ω Z3

1F i

Z1 1Ω ∼

Fig. 3.68 A circuit with voltage sources

cos(t)A

∼ + cos(t)V Z2 v Z1

2Ω Z3 i 3Ω

1F

sin(t)V +∼

4

Steady-State Power

Electric power is the rate at which electric energy is transferred from an electric power source to a sink that absorbs power. For example, a bulb emits light by absorbing electric power. All the three entities, voltage, current, and power, are important in the generation, transmission, distribution, and utilization of electric power. All the devices, such as a generator, a motor, a heater, and an amplifier, are characterized by their ability to generate power or consume power to do some work. Circuit element resistor dissipates energy. However, capacitors and inductors store energy, respectively, in their electric and magnetic fields, and, hence, are called storage elements. The stored energy in inductors and capacitors can sustain current in the circuit for some time after the power is switched off. That is why we get light, although diminishing, from the bulb for some time after the power is switched off. While the expression for the power is still of the form V I , the power in AC circuits has two components to take into account, the active power and reactive power. The stored energy is returned to the source. Therefore, the average power consumed by the storage devices, in the sinusoidal steady state, is zero. The kinetic energy is stored in the magnetic field generated by the flow of current in the inductor. Similar to the kinetic energy, 12 Mv 2 in a mechanical system, the energy stored in an inductor of L henries is 12 Li 2 (t). The current i(t) flowing through an inductor generates a voltage L di(t) dt . Therefore, power has to be delivered by the source deriving a current through the inductor to oppose the generated voltage. This power is used to store energy in the inductor. Therefore, energy delivered, dT (t), during an infinitesimal interval dt is dT (t) = v(t)i(t)dt = i(t)L

di(t) dt. dt

Integrating this expression, we get the expression for the instantaneous energy stored in the inductor as  T (t) = L

t

−∞

i(x)

1 di(x) dx = Li 2 (t), dx 2

since i(−∞) = 0. A capacitor is made of two conducting surfaces separated by a dielectric (nonconductor or insulator) material. When we apply a voltage v(t) across a capacitor, an electric field is set up. Therefore, power has to be delivered by the source. This power is used to store energy in the capacitor. Therefore, energy delivered, dV (t), during an interval dt is © Springer Nature Switzerland AG 2020 D. Sundararajan, Introductory Circuit Theory, https://doi.org/10.1007/978-3-030-31985-4_4

149

150

4 Steady-State Power

dV (t) = v(t)i(t)dt = v(t)C

dv(t) dt. dt

Integrating this expression, we get the expression for the instantaneous energy stored in the capacitor as  t 1 dv(x) dx = Cv 2 (t) V (t) = C v(x) dx 2 −∞ The current i(t) and the voltage v(t) may be any functions of time and T (t) and V (t) are the corresponding time functions representing the stored energies at instants of time. Work has to be done to move a charge in a certain direction in a conductor. The work is carried out by the voltage between the two points of interest. Therefore, v=

dw , dq

where w is the work in joules, v is the voltage in volts, and q is the charge in coulombs. That is, one volt is one joule per coulomb. Now, power p is the rate of doing work measured in watts. p=

dw dw dq = = vi. dt dq dt

If p is positive, power is absorbed by the element. Otherwise, power is generated by the element. Energy measured in joules, is the integral of power over an interval. A resistor always absorbs power. The average power absorbed by inductors and capacitors is zero. The power supplied by the sources is equal to the power dissipated in the resistors and that stored in the reactive elements.

4.1

Energy in Reactive Elements with Sinusoidal Sources

Let the source to a capacitor C be v(t) = V cos(ωt). The instantaneous energy stored in its electric field associated with the capacitor is V (t) =

1 1 CV 2 cos2 (ωt) = CV 2 (1 + cos(2ωt)). 2 4

We used the trigonometric identity cos2 (ωt) =

1 (1 + cos(2ωt)). 2

This expression represents a shifted double-frequency sinusoid with its amplitude varying from 0 to 0.5CV 2 . The average energy is 1 Vav = CV 2 . 4 Differentiating the expression for V (t) with respect to t, we get dV (t) ω = − CV 2 sin(2ωt). dt 2

4.2 Power Relations in a Circuit

151

Therefore, dV (t) |av = 0, dt since the integral of a sinusoid over an integral number of its cycles is zero. Let the source to an inductor L be i(t) = I cos(ωt). The instantaneous energy stored in its magnetic field associated with the inductor is T (t) =

1 2 1 LI cos2 (ωt) = LI 2 (1 + cos(2ωt)). 2 4

The average energy is 1 2 LI . 4 Differentiating the expression for T (t) with respect to t, we get Tav =

dT (t) ω = − LI 2 cos(2ωt). dt 2 Therefore, dT (t) |av = 0. dt A reactive element cannot absorb energy indefinitely.

4.2

Power Relations in a Circuit

Let the voltage applied to a circuit be v(t) = V cos(ωt + θ ). Let the current through the circuit be i(t) = I cos(ωt + φ). The instantaneous power absorbed by the circuit is s(t) = v(t)i(t) = V I cos(ωt + θ ) cos(ωt + φ). Using the trigonometric identity 2 cos(A) cos(B) = cos(A − B) + cos(A + B), we get s(t) = 0.5V I cos(θ − φ) + 0.5V I cos(2ωt + θ + φ). The first of the two components of s(t) is a constant and its value depends on the difference between the phases of the voltage and current. The second component is a sinusoid of double the frequency of that of the source. Since s(t) is time varying, the average value S of s(t) over one period, with unit watts, is measured by the wattmeter, the instrument to measure the average power. Let the period of the source be 2T . Since the period of s(t) is T , the apparent average power, with unit volt amperes, is given by 1 S= T

 0

T

1 0.5V I cos(θ − φ)dt + T



T

0.5V I cos(2ωt + θ + φ)dt = 0.5V I cos(θ − φ).

0

The second integral involves a sinusoid and the integral of a sinusoid over its period is zero.

152

4 Steady-State Power

In the frequency-domain representation of circuits, the voltages and currents are, respectively, represented as V  θ and I  φ. Then, S = 0.5(V  θ )(I  (−φ)) = 0.5V I  (θ − φ) = 0.5V I (cos(θ − φ) + j sin(θ − φ)) = Pav + j Qav . For passive circuits, S will lie in the first or fourth quadrants of the complex plane. If S lies in the first quadrant, the load is inductive. If S lies in the fourth quadrant, the load is capacitive. Taking the magnitude, we get  |S| =

2 + Q2 , Pav av

S is the apparent power. The first component, which is the power actually consumed, is called the active power Pav . The power that is swapped back and forth from the source and the circuit is called the reactive or wattless power Qav . Therefore, Pav = 0.5Re((V  θ )(I  (−φ))) = 0.5V I cos(θ − φ) Qav = 0.5Im((V  θ )(I  (−φ))) = 0.5V I sin(θ − φ). That is the real part of the 0.5 times the product of the complex amplitude of the voltage and the conjugated complex amplitude of the current is the active power. The imaginary part is the reactive power. The active power is usually denoted by the symbol Pav and the reactive power by the symbol Qav . Therefore, the power in complex form is S = 0.5V I ∗ = 0.5I I ∗ Z = 0.5|I |2 Z = 0.5|V |2 Y ∗ = Pav + j Qav , where Z is the impedance, Y = 1/Z is the admittance, and V = V θ

and

I = I  φ.

The active power in a linear passive circuit is always positive, while the reactive power is negative in capacitive circuits and positive in inductive circuits. The unit for reactive power is vars (volt–amperes reactive). The total power consumed by a circuit is the sum of the powers consumed by the constituent branches of the circuit. Since the reactive power may be positive or negative, it is possible to change its value by adding suitable passive circuits, a process called power-factor correction. It should be mentioned that √ the meters used to measure currents and voltages measure the RMS values, which is equal to 1/ 2 of the peak value of a sinusoid. If we use RMS values, then the expressions for the power become VRMS IRMS , as in the case of DC circuits without the term 1/2. If θ = φ, as in resistive circuits with the phase difference zero, the average power is P = 0.5 VI. That is, the effective impedance has only resistive component. If θ −φ = ±0.5π , the average power is zero. That is, the effective impedance has only reactive component. In the term cos(θ − φ), the phase difference is usually represented by θ . As cos(θ ) indicates the extent the circuit absorbs power from the source, it is called the power factor of the circuit, denoted by pf. The cosine of the phase difference between the voltage and current is the power factor. It is also the cosine of the angle of the impedance.

4.2 Power Relations in a Circuit

153

A capacitive impedance has a negative phase angle. Consequently, the reactive power Qav , which is a function of the angle, is negative and vice versa for an inductive impedance. The pf is an indication of the relative magnitudes of the resistive and reactive components of a circuit. A higher pf indicates that the resistive component is greater. The total average power supplied is the sum of Pav for each element of the circuit. Pav = Pav1 + Pav2 + Pav3 + · · · Similarly, the total reactive power supplied is the algebraic sum of Qav for each element of the circuit. Qav = Qav1 + Qav2 + Qav3 + · · · Therefore, Sav = Sav1 + Sav2 + Sav3 + · · · The equivalent definitions of the power factor are pf =

active power R = cos(θ ) = . |apparent power| |Z|

Angle θ is the phase angle of the current with respect to the voltage. If the current leads the voltage, the power factor is said to be leading and is lagging if current lags voltage. A pf of 0.8 lagging implies that the current lags the voltage by cos−1 (0.8) = 0.6435 radians or 36.8699◦ . 1. For resistive loads, Q = 0 and pf = 1. 2. For inductive loads, Q > 0 with a lagging pf. 3. For capacitive loads, Q < 0 with a leading pf. Example 4.1 Find the average active and reactive power consumed by the circuit, shown in Fig. 4.1a. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Relate the stored energy with the reactive power. Solution The source voltage is 0.5 − j 0.866 V. The frequency of the source is ω = 2 radians. The values of the impedances are Z1 = 1, Z2 = j 4. 0.866 The source current is I = 0.5−j = −0.1744 − j 0.1686 A. 1+j 4 The values of the currents through the impedances are

1Ω (a)

+ ∼

1F

cos(2t)V

2H + ∼

cos(t+ π6 )V

cos(2t− π3 )V

{IZ1 = −0.1744 − j 0.1686, IZ2 = −0.1744 − j 0.1686}.

2H + ∼

0.2F

1Ω

3Ω

(b)

(c)

Fig. 4.1 (a) A RL series circuit; (b) a RC series circuit; (c) a RLC series circuit

154

4 Steady-State Power

The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are 0.0294, j 0.1176. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.0294 + j 0.1176 va. The power supplied by the source V I ∗ /2 is also the same. The average power consumed by the resistor is 0.0294 W and is equal to Pav . The impedance of the circuit is Z = V /I = 1 + j 4. pf =

Re(Z) 0.0294 = = cos( (V ) −  (I )) = 0.2425 lagging. |Z| |S|

The instantaneous energy dissipated by the resistor is given, in the time-domain, by p(t) = i 2 (t)R. For this example, R = 1 and i(t) = |I | cos(2t +  (I ))A = 0.2425 cos(2t − 2.373). Using the trigonometric double-angle formula cos2 (x) =

1 + cos(2x) , 2

we can find p(t) = i 2 (t) to be 0.24252 cos2 (2t − 2.373) = 0.0294 + 0.0294 cos(4t + 1.5372).

0.0588

0.0294

0.0294

0

P

P

The period of this waveform is 2π/(4) = π/2 seconds. Alternatively, p(t) can also be computed numerically at discrete points with sufficiently small interval to approximate the instantaneous power. The result is shown in Fig. 4.2a. It oscillates about the average power 0.0294. The phase is about 90◦ .

0.00000 0

1.5708

t

(a)

-0.0294 0

1.5708

t

(b)

Fig. 4.2 (a) Instantaneous energy dissipated by the resistor; (b) the oscillating component of the energy

4.2 Power Relations in a Circuit

155

That is, it is an inverted sine wave. The oscillating part alone is shown in Fig. 4.2b. The oscillating part is a double-frequency sinusoid. Manually, one can do the computation at few points and find that the power waveform has two peaks and two valleys, in a period of the current through the resistor, with all the values positive. Let the time-domain sinusoidal current be expressed in complex form. Then, ip (t) =

 1  j ωt Ip e + Ip∗ e−j ωt . 2

The product of two currents ip (t) and iq (t) is   1  j ωt Ip e + Ip∗ e−j ωt Iq ej ωt + Iq∗ e−j ωt 4  1 = Ip Iq ej 2ωt + Ip∗ Iq∗ e−j 2ωt + Ip Iq∗ + Ip∗ Iq 4  1  ∗ 1  = Ip Iq + Re Ip Iq ej 2ωt . 2 2

ip (t)iq (t) =

Since ip (t) = iq (t) = i(t) and the energy dissipated in a resistor is Ri 2 (t), we get the instantaneous power dissipated as P (t) =

R 2 R  2 j 2ωt  |I | + Re I e . 2 2

The formula i 2 R is for a DC current. Since the amplitude profile of a sinusoidal waveform is time varying, it is found that Ieff = IRMS = 0.7071|I | serves the same purpose in power calculations, where |I | is the peak value. The constant value Ieff dissipates the same power in a resistor as that by a sinusoidal current with a peak value of |I | on the average. That is the average power dissipated in a resistor due to a sinusoidal current flowing through it is |I | |I | |I |2 2 R=√ √ R R= Ieff 2 2 2 |I | √ | are referred as which is the first part of the expression for power derived above. The values √ and |V 2 2 the effective values of sinusoidal current or voltage. They are also known as RMS (root-mean-square) values by their definition. A similar expression for the instantaneous energy stored in a magnetic field is

T (t) =

L 2 L  2 j 2ωt  |I | + Re I e . 4 4

The first part is the average energy stored. As the energy stored is 0.5Li 2 (t) rather than Ri 2 (t), L replaces R and the denominator is multiplied by 2. For this example, as L = 2 and R = 1 and with the same current, we get the same figure, Fig. 4.3, for the stored energy.

156

4 Steady-State Power

T

0.0588

0.0294

0.0000 0

1.5708

t Fig. 4.3 Instantaneous energy stored in the inductor

As expected, the stored energy is related to the reactive power by a constant factor 2ω. The equilibrium equation for the circuit, in the frequency-domain with v(t) → V ej ωt is

and

i(t) → I ej ωt ,

I R + j ωLI = V .

Multiplying both sides by 0.5I ∗ , we get 0.5I I ∗ R + j 2ωL0.25(I I ∗ ) = 0.5V I ∗ = Pav + j Qav . Therefore, for this example, the average reactive power is 2(2)0.0294 = 0.1176 

as obtained above.

Example 4.2 Find the average active and reactive power consumed by the circuit, shown in Fig. 4.1b. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Relate the stored energy with the reactive power. Solution The source voltage is 0.8660 + j 0.5 V. The source frequency is ω = 1 radian. The values of the impedances are Z1 = 1, Z2 = −j 1. 0.5 The source current is I = 0.8660+j = 0.1830 + j 0.6830 A = 0.7071 1.3090. The values of the 1−j 1 currents through the impedances are

{IZ1 = 0.1830 + j 0.6830, IZ2 = 0.1830 + j 0.6830}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are 0.25, −j 0.25. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.25 − j 0.25 va.

157

0.5000

0.2500

0.25

0

P

P

4.2 Power Relations in a Circuit

0.0335 0

-0.2500 0

3.1416

t

3.1416

t

Fig. 4.4 (a) Instantaneous energy dissipated by the resistor; (b) the oscillating component of the energy

The power supplied by the source V I ∗ /2 is also the same. The average power consumed by the resistor is 0.25 W and is equal to Pav . The impedance of the circuit is Z = V /I = 1 − j . pf =

Re(Z) 0.25 = = cos( (V ) −  (I )) = 0.7071 leading. |Z| |S|

The instantaneous energy dissipated by the resistor is 0.70712 cos2 (t + 1.3090) = 0.25 + 0.25 cos(2t + 2.618). The period of this waveform is 2π/(2) = π seconds. The result is shown in Fig. 4.4a. It oscillates about the average power 0.25. The phase is about 150◦ . That is, it is an inverted and shifted sine wave. The oscillating part alone is shown in Fig. 4.4b. An expression, similar to that for magnetic energy, for the instantaneous energy stored in an electric field is   1 1 |I |2 − Re I 2 ej 2ωt , V (t) = 2 2 4Cω 4Cω where I is the current through the capacitor and C is the value of the capacitor. Note that V =

I . j ωC

The first part is the average energy stored. The equilibrium equation for the circuit, in the frequency-domain with v(t) → V ej ωt is

and

IR − j

Multiplying both sides by 0.5I ∗ , we get

i(t) → I ej ωt ,

I = V. ωC

  (I I ∗ ) = 0.5V I ∗ = Pav + j Qav . 0.5I I ∗ R − j 2ω 0.25 Cω2

158

4 Steady-State Power

V

0.2500

0.125

0.0000 0

3.1416

t Fig. 4.5 Instantaneous energy stored in the capacitor

The reactive power is −2ω times the energy stored in the capacitor. Therefore, for this example, the average reactive power, with the squared magnitude of the current being 0.5, is −2(1)0.125 = −0.25 as obtained above. The instantaneous energy stored in the electric field is shown in Fig. 4.5.



Example 4.3 Find the average active and reactive power consumed by the circuit, shown in Fig. 4.1c. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Relate the stored energy with the reactive power. Solution The source voltage is 1 V. The frequency of the source is ω = 2 radians. The values of the impedances are Z1 = 3, Z2 = j 4, Z3 = −j 2.5. The source current is I=

1 = (0.2667 − j 0.1333) A. 3 + j 1.5

The values of the currents through the impedances are {IZ1 = 0.2667 − j 0.1333, IZ2 = 0.2667 − j 0.1333, IZ3 = 0.2667 − j 0.1333}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are 0.1333, j 0.1778, −j 0.1111. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.1333 + j 0.0667 va. The power supplied by the source V I ∗ /2 is also the same. The average power consumed by the resistor is 0.1333 W and is equal to Pav . The impedance of the circuit is Z = V /I = 3 + j 1.5.

159

0.2667

0.0889

0.1333

0.0444

T

P

4.3 Power-Factor Correction

0.00000 0

t

(a)

0.0556

0.3555

0.0278

0.0677 0

Q

V

0.0000 0

1.5708

0.0000 0

1.5708

t

1.5708

t

(b)

t

(d)

-0.2222 0

(c)

1.5708

Fig. 4.6 (a) Instantaneous energy dissipated by the resistor; (b) instantaneous energy stored in the inductor; (c) instantaneous energy stored in the capacitor; (d) reactive energy component of the circuit

pf =

0.1333 Re(Z) = = cos( (V ) −  (I )) = 0.8944. |Z| |S|

The instantaneous energy dissipated by the resistor is (3)0.29812 cos2 (2t − 0.4636) = 0.1333 + 0.1333 cos(4t − 0.9273). The period of this waveform is 2π/(4) = π/2 seconds. The result is shown in Fig. 4.6a. It oscillates about the average energy 0.1333. The phase is about −53.1301◦ . The instantaneous energy stored in the magnetic field is shown in Fig. 4.6b. The instantaneous energy stored in the electric field is shown in Fig. 4.6c. Figure 4.6d shows the reactive energy consumed by the circuit, which is obtained by multiplying the difference in the stored energies by the inductor and capacitor multiplied by 2ω. The lower limit is negative indicating that the source receives part of the reactive energy from the storage elements. 

4.3

Power-Factor Correction

Most of the loads, such as induction motors and furnaces, are inductive. The power has two components: (1) active power to do the work such as heating, producing light, motion, etc.; and (2) the reactive power to sustain the magnetic field associated with such loads. With a pf of 0.8, 100 va is required to get 80 watts of power. As the pf decreases, a higher current is required from the supply line to do the same work. If the used power is charged on va basis, obviously, pf correction saves money. If the

160

4 Steady-State Power

used power is charged on watts basis, then no additional charges are made for pf between some limits, say 0.9 to 1. If the pf goes lower than 0.9, then additional charges are made as per the agreed contract. For inductive loads with a voltage source, a suitable parallel capacitor C has to be connected to improve the power factor. With a fixed terminal voltage, the current to the net impedance will be reduced. The active power remains the same, while the reactive power gets reduced resulting in improved pf. To improve the pf, capacitors can be installed at the load itself or at the feeder or substation. Let the reactive power be Q1 and the desired one be Q2 . Since, Q1 − Q2 = 0.5

V2 = 0.5ωCV 2 , XC

Q1 − Q2 . ωV 2 Although most practical loads are inductive, it is possible to improve the power factor for capacitive loads also in a similar manner by connecting an inductor L in parallel. Since, C=2

Q1 − Q2 = 0.5 L=

V2 V2 , = 0.5 XL ωL

V2 . 2ω (Q1 − Q2 )

With a current source, an appropriate impedance has to be connected in series with the load impedance. Consider a circuit, shown in Fig. 4.7a, with the voltage applied being  π . v(t) = cos 2t + 6 The impedance of the circuit is composed of a resistor 1 connected in series with a capacitor 0.5 F. Find the apparent, active, and reactive powers and the pf of the circuit. If the pf is to be 0.8 and 1, what is the value of the inductor to be connected in parallel with the impedance in each case? The frequency-domain representation of the voltage and the impedance is π = 1 30◦ = 0.8660 + j 0.5 6 pf =

+ ∼

0.5F 1Ω (a)

Z = 1 − j1

R = 0.7071 |Z| pf = 0.8

4H

cos(2t+ π6 )V

cos(2t+ π6 )V

pf = 0.7071

and

+ ∼

0.5F 1Ω (b)

pf = 1

1H

cos(2t+ π6 )V

V = 1

+ ∼

0.5F 1Ω (c)

Fig. 4.7 (a) A RC series circuit; (b) the circuit with a pf correction inductor of 4 H ; (c) the circuit with a pf correction inductor of 1 H

4.3 Power-Factor Correction

161

leading. The current through the circuit is I=

V = 0.1830 + j 0.6830 = 0.7071 75◦ Z

S = 0.5V I ∗ = 0.5(0.8660 + j 0.5)(0.1830 − j 0.6830) = 0.25 − j 0.25 = Pav + j Qav pf =

Pav = 0.7071 |S|

as obtained above. The power consumed by the resistor is 0.5I ∗ I (1) = 0.25W. Figure 4.8a shows the voltage and the current. The current is leading the voltage by 45◦ . Figure 4.8d shows the active and reactive power components. The desired pf is 0.8. As the circuit is capacitive, an inductor has to be connected in parallel with the impedance to improve the pf from 0.7071 to 0.8. The active power remains the same, but the reactive power drawn from the source has to be reduced. The balance is supplied by the inductor, as shown in Fig. 4.7b. In the steady state, part of the reactive power is supplied by the inductor. That is, energy stored in the magnetic field associated with the inductor is swapped back and forth with the electric field associated with the capacitor, reducing the amount of reactive power to be supplied by the voltage source. The new apparent power is Pav = 0.3125 va. 0.8 Therefore, the new reactive power is   0.3125 sin cos−1 (0.8) = 0.1875vars.

0.5

0.5748 0.5 V

0 0 0.183

I

imag

imag

0.5

0 I -0.1083 L 0.06 0.25

0.866

real (a)

Q av

Q av -0.25 0

P av

0.25 (d)

0.866

0

-0.1875 0

P av

0.25 (e)

I

0

-0.43

real (b)

0

V

0.25

V

imag

I

IL 0.25 0.43

0.866

real (c)

Q av

0.683

0

0

P av

0.25 (f)

Fig. 4.8 (a) V and I in Fig. 4.7a; (b) V and I in Fig. 4.7b; (c) V and I in Fig. 4.7c; (d) apparent power in Fig. 4.7a; (e) apparent power in Fig. 4.7b; (f) apparent power in Fig. 4.7c

162

4 Steady-State Power

The difference between the reactive powers is 0.25−0.1875 = 0.0625. Now, with ω = 2 and |V | = 1, L=

12 = 4H. (2)(2)0.0625

The current through the inductor is V = 0.0625 − j 0.1083. j 2(4) The new current supplied by the source is I n = (0.1830 + j 0.6830) + (0.0625 − j 0.1083) = 0.2455 + j 0.5748. The power supplied is S = 0.5V I n∗ = 0.5(0.8660 + j 0.5)(0.2455 − j 0.5748) = 0.2500 − j 0.1875 = Pav + j Qav . The new pf is pf =

Pav = 0.8. |S|

Figure 4.8b shows the voltage, currents from the source and the inductor. The phase difference between the voltage and the current has been reduced, improving the pf (which is the cosine of this phase angle). Figure 4.8e shows the active and reactive power components. The desired pf is 1. The whole reactive power 0.25 has to be supplied by the inductor. Obviously, the inductor has to supply more current and it requires a smaller inductor, as shown in Fig. 4.7c. L=

12 = 1H. (2)(2)0.25

The current through the inductor is V = 0.2500 − j 0.4330. j 2(1) The new current supplied by the source is I n = (0.1830 + j 0.6830) + (0.2500 − j 0.4330) = 0.4330 + j 0.25. The power supplied is S = 0.5V I n∗ = 0.5(0.8660 + j 0.5)(0.4330 − j 0.25) = 0.2500 − j 0 = Pav + j Qav . The new pf is pf =

Pav = 1. |S|

Figure 4.8c shows the voltage, currents from the source and the inductor. The phase difference between the voltage and the current is zero, making the pf equal to 1. Figure 4.8f shows the active and reactive power components. Figure 4.9 shows the instantaneous power consumed in the circuit with different pf.

4.3 Power-Factor Correction

163

0.6035 0.5625 0.5000

pf=0.7071

p(t)

pf=1

0.25

pf=0.8

0.0000 -0.0625 -0.1036 0

1

1.309

t Fig. 4.9 The instantaneous power consumed by the circuit is the sum of a constant and a double-frequency sinusoidal component

Example 4.4 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.13a. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Find the value of the capacitor for power-factor improvement, for pf = 0.8 lagging and pf = 1. Solution The source voltage is 0.866 + j 0.5 V. The source current is I = I1 = 0.6550 − j 0.1845 A. The values of the impedances are Z1 = −j 10, Z2 =

j3 −j 10 = 0.3 + j 0.9, Z3 = j 1, Z4 = = 0.9901 − j 0.0990, Z5 = 1. 3 + j1 1 − j 10

The values of the currents through the impedances are {IZ1 = −0.0283 + j 0.0677, IZ2 = 0.6833 − j 0.2523, IZ3 = 0.2171 − j 0.1886, IZ4 = 0.4379 + j 0.0041, IZ5 = −0.2454 + j 0.2564}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are −j 0.0269, 0.0796 + j 0.2387, j 0.0414, 0.0950 − j 0.0095, 0.0630. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.2375 + j 0.2437 va. The power supplied by the source V I ∗ /2 is also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 3, 1, 1 and the corresponding currents are found from the branch currents, respectively, as 0.1440 + j 0.1798,

0.4340 − j 0.0393

− 0.2454 + j 0.2564.

164

4 Steady-State Power

The total power consumed by the resistors is 0.0796 + 0.0950 + 0.063 = 0.2375 W and is equal to Pav found earlier. The impedance of the circuit is Z = V /I = 1.0257 + j 1.0523. pf =

Re(Z) Pav = = cos( (V ) −  (I )) = 0.698. |Z| |S|

The desired power factor is 0.8. The corresponding apparent power is Pav = 0.2969. 0.8 The corresponding reactive power is 0.2969 sin(cos−1 (0.8)) = 0.1781. The difference between the reactive powers is 0.2437 − 0.1781 = 0.0655. Now, with ω = 1 and |V | = 1, 0.0655 = 0.1311F. C=2 (1)(12 ) The current through the capacitor is j V (1)(0.1311) = −0.0655 + j 0.1135. The new current supplied by the source is (0.6550 − j 0.1845) + (−0.0655 + j 0.1135) = 0.5895 − j 0.0710. The new pf is cos

π

 −  (0.5895 − j 0.0710) = 0.8.

6 The desired power factor is 1. The corresponding apparent power is Pav = 0.2375. 1 The corresponding reactive power is   0.2375 sin cos−1 (1) = 0.

The difference between the reactive powers is 0.2437 − 0 = 0.2437. Now, with ω = 1 and |V | = 1, C=2

0.2437 = 0.4873F. (1)(12 )

4.3 Power-Factor Correction

165

The current through the capacitor is j V (1)(0.4873) = −0.2437 + j 0.4220. The new current supplied by the source is (0.6550 − j 0.1845) + (−0.2437 + j 0.4220) = 0.4114 + j 0.2375. The new pf is cos

π 6

 −  (0.4114 + j 0.2375) = 1. 

Example 4.5 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.14a. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Find the value of the capacitor for power-factor improvement, for pf = 0.8 lagging. Solution The source voltage is 0.5 + j 0.866 V. The source current is I = Iz5 = −0.4192 − j 0.0678 A. The values of the impedances are Z1 = j 1, Z4 =

Z2 =

j2 = 0.4 + j 0.8, 2 + j1

−j 30 = 2.7523 − j 0.8257, 3 − j 10

Z3 = −j 10,

Z5 = 1,

Z6 = 1 + j 1.

The values of the corresponding currents are {IZ1 = −0.4837 − j 0.0727, IZ2 = 0.3912 + j 0.0038, IZ3 = −0.0645 − j 0.0049, IZ4 = −0.0280 − j 0.0640, IZ5 = −0.4192 − j 0.0678, IZ6 = −0.0925 − j 0.0689}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are j 0.1196, 0.0306 + j 0.0612, −j 0.0209, 0.0067 − j 0.0020, 0.0902, 0.0067 + j 0.0067. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.1342 + j 0.1646 va. The power supplied by the source V I ∗ /2 is also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 2, 3, 1, 1 and the corresponding currents are found from the branch currents, respectively, as 0.0767 − j 0.1572,

−0.0433 + j 0.0510

− 0.4192 + j 0.0678,

−0.0925 + j 0.0689.

166

4 Steady-State Power

The total power consumed by the resistors is 0.0306 + 0.0067 + 0.0902 + 0.0067 = 0.1342 W and is equal to Pav found earlier. The impedance of the circuit is Z = V /I = 1.4878 + j 1.8252. pf =

Re(Z) Pav = = cos( (V ) −  (I )) = 0.6318. |Z| |S|

The desired power factor is 0.8. The corresponding apparent power is Pav = 0.1677. 0.8 The corresponding reactive power is   0.2969 sin cos−1 (0.8) = 0.1006. The difference between the reactive powers is 0.1646 − 0.1006 = 0.0640. Now, with ω = 1 and |V | = 1, 0.0640   = 0.1279F. C=2 (1) 12 The current through the capacitor is j V (1)(0.1279) = −0.1108 + j 0.0640. The new current supplied by the source is (0.4192 + j 0.0678) + (−0.1108 + j 0.0640) = 0.3084 + j 0.1318. The new pf is cos

π 3

 −  (0.3084 + j 0.1318) = 0.8. 

Example 4.6 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.15a. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Find the value of the capacitor for power-factor improvement, for pf= 0.9 lagging. Solution The source voltage is 0.5 + j 0.8660 V. The source current is I = −0.5112 + j 0.1373 A. The values of the impedances are Z1 = j 1, Z4 =

Z2 =

j2 = 0.4 + j 0.8, 2 + j1

−j 30 = 2.7523 − j 0.8257, 3 − j 10

Z3 = −j 10, Z5 = 1 + j 1.

4.3 Power-Factor Correction

167

The values of the corresponding currents are {IZ1 = −0.4837 − j 0.0727, IZ2 = 0.3912 + j 0.0038, IZ3 = −0.0645 − j 0.0049, IZ4 = −0.0280 − j 0.0640, IZ5 = −0.4192 − j 0.0678, IZ6 = −0.0925 − j 0.0689}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are j 0.1859, 0.0476 + j 0.0951, −j 0.0325, 0.0104 − j 0.0031, 0.0103 + j 0.0103. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.0683 + j 0.2557 va. The power supplied by the source V I ∗ /2 is also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 2, 3, 1 and the corresponding currents are found from the branch currents, respectively, as 0.1676 + j 0.1395,

−0.0753 − j 0.0359

− 0.1404 − j 0.0310.

The total power consumed by the resistors is 0.0476 + 0.0104 + 0.0103 = 0.0683 W and is equal to Pav found earlier. The impedance of the circuit is Z = V /I = 0.4878 + j 1.8252. pf =

Re(Z) Pav = = cos( (V ) −  (I )) = 0.2582. |Z| |S|

The desired power factor is 0.9. The corresponding apparent power is Pav = 0.0759. 0.9 The corresponding reactive power is   0.0759 sin cos−1 (0.9) = 0.0331. The difference between the reactive powers is 0.2557 − 0.0331 = 0.2226. Now, with ω = 1 and |V | = 1, 0.2226 = 0.4452F. C=2 (1)(12 ) The current through the capacitor is j V (0.4452) = −0.3855 + j 0.2226.

168

4 Steady-State Power

The new current supplied by the source is (0.5112 − j 0.1373) + (−0.3855 + 0.2226) = 0.1257 + j 0.0853. The new pf is cos

π 3

 −  (0.1257 + j 0.0853) = 0.9. 

Example 4.7 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.16a. What is the power factor of the circuit? Solution The source voltages are −j and 2 V. The source current is −j A. The values of the impedances are Z1 = j 0.1,

Z2 = 2 − j 5,

Z3 = 2,

Z4 = −j 10,

Z5 = 1 + j 0.2.

The values of the corresponding currents are {IZ1 = 0.1821 + j 0.9461, IZ2 = 0.1821 − j 0.0539, IZ3 = j 1, IZ4 = −0.0824 + j 0.1921, IZ5 = 0.0824 + j 0.8079}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2, are j 0.0464, 0.0361 − j 0.0901, 1, −j 0.2185, 0.3298 + j 0.0660. The total apparent power consumed by the circuit is the sum of those of all the impedances S = Pav + j Qav = 1.3658 − j 0.1962 va. The power supplied by the sources is (−0.4730 − j 0.0910) + (1.9213 + j 0.0869) + (−0.0824 − j 0.1921) also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 2, 2, 1 and the corresponding currents are found from the branch currents, respectively, as 0.1821 − j 0.0539,

1,

0.0824 + j 0.8079.

The total power consumed by the resistors is 0.0361 + 1 + 0.3298 = 1.3658 W and is equal to Pav found earlier.

4.3 Power-Factor Correction

169

The power factor of the circuit is pf =

Pav = 0.9898 |S|

leading. As it is close to 1, power-factor correction may not be needed.  Example 4.8 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.17a. What is the power factor of the circuit? Solution The source voltage is −j . The source current is 2 A. The values of the impedances are Z1 = j 0.1,

Z2 = 0.9615 − j 0.1923,

Z3 = 2.

The values of the corresponding currents are {IZ1 = 1.9701 − j 0.5985, IZ2 = 2, IZ3 = −0.0299 − j 0.5985}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2, are j 0.2120, 1.9231 − j 0.3846, 0.3591. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 2.2822 − j 0.1726 va. The power supplied by the sources is (0.2993 − j 0.9850) + (1.9830 + j 0.8124) also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 1, 2 and the corresponding currents are found from the branch currents, respectively, as 1.9231 − j 0.3846,

−0.0299 − j 0.5985.

The total power consumed by the resistors is 1.9231 + 0.3591 = 2.2822 W and is equal to Pav found earlier. The power factor of the circuit is pf =

Pav = 0.9972 |S|

leading. As it is close to one, power-factor correction may not be needed.



170

4 Steady-State Power

If the current enters the positive terminal of the source, the source is absorbing power. Otherwise, the source is supplying power. The total average power is equal to the sum of the powers supplied by each source acting alone, when the frequencies are harmonic. That is, the higher frequencies are integral multiple of the lowest nonzero frequency. Example 4.9 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.18. What is the power factor of the circuit? Solution The DC source is 2 V. The AC source voltage is 0.8660 − j 0.5 V at ω = 1 radian. The source current is 2 A at ω = 5 radians. The impedances are Z1 = j ω0.1, Z2 =

1 (1/(j ω0.1))2 , Z3 = 10, Z4 = 2, Z5 = , Z6 = j ω0.2, Z7 = 1. (1 + j ω0.2) (2 + (1/(j ω0.1))

With the DC source alone, the resistor values are 1 and 2 in series. The current is 2/3 A. Therefore, the power consumed is 22 4 2 3 = = 2 W. 33 3 3 The pf is equal to 1. With ω = 1, the values of the impedances are Z1 = j 0.1, The value of the current is

Z2 = 0.9615 − j 0.1923,

Z3 = 10.

I = 0.0794 − j 0.0449.

The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are j 0.0004, 0.0040 − j 0.0008, 0.0416. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 0.0456 − j 0.0004 va. The power supplied by the source is 0.0456 − j 0.0004 also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 10, 1 and the corresponding currents are found from the branch currents, respectively, as 0.0794 − j 0.0449,

0.0677 − j 0.0585.

The total power consumed by the resistors is 0.0416 + 0.0040 = 0.0456 W and is equal to Pav found earlier.

4.3 Power-Factor Correction

171

The power factor of the circuit is pf =

Pav = 0.999965 |S|

leading. As it is close to 1, power-factor correction may not be needed. With ω = 5, the values of the impedances are Z1 = j 0.5, Z4 = 2,

Z2 = 0.5 − j 0.5,

Z5 = 1 − j 1,

Z3 = 10

Z6 = j 1,

Z7 = 1.

The value of the currents, respectively, are 2 − j 0.0952, j 0.0952, j 0.0952, 2, 1 + j 1, 1 − j 1, 1 − j 1. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are j 1.0023, 0.0023 − j 0.0023, 0.0454, 4, 1 − j 1, j 1, 1. The total apparent power consumed by the circuit is the sum of those of all the impedances S = Pav + j Qav = 6.0476 + j 1 va. The power supplied by the source is also the same. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 1, 10, 2, 2, 1 and the corresponding currents are found from the branch currents, respectively, as 0.0476 + j 0.0476, j 0.0952, 2, 1, 1 − j 1. The total power consumed by the resistors is 0.0023 + 0.0454 + 4 + 1 + 1 = 6.0476 W and is equal to Pav found earlier. The power factor of the circuit is pf = lagging.

Pav = 0.9866 |S| 

Example 4.10 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.19. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components.

172

4 Steady-State Power

Solution The source voltage is V = 2 V. The source current is I = −j A. The values of the impedances are Z1 = 2 + j 0.1,

Z2 = 1,

Z3 = 3 − j 10,

Z4 = 3,

Z5 = 2.

The values of the corresponding currents are {IZ1 = 1.7257 + j 0.7736, IZ2 = −5.4514 + j 0.4529, IZ3 = 0.3666 + j 0.4978, IZ4 = 1.3591 − j 0.7242, IZ5 = −3.7257 + j 0.2264}. The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are 3.5765 + j 0.1788, 14.9615, 0.5732 − j 1.9107, 3.5575, 13.9321. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 36.6008 − j 1.7319 va. The power supplied by the sources is also the same. (1.3127 + j 5.7644) + (−5.0848 − j 0.9506) + (40.3729 − j 6.5457) = 36.6008 − j 1.7319. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 2, 1, 3, 3, 2 and the corresponding currents are the same as the branch currents. The total power consumed by the resistors is 3.5765 + 14.9615 + 0.5732 + 3.5575 + 13.9321 = 36.6008 W and is equal to Pav found earlier. pf =

Pav = 0.9989 |S| 

leading.

Example 4.11 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.20. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Solution The source voltage is j V. The values of the impedances are Z1 = 1 + j 0.1,

Z2 = 3,

Z3 = 0.0099 + j 0.0990,

Z4 = 3,

Z5 = 1.

The values of the corresponding currents, respectively, are 0.9472 + j 2.6635, −0.2664 − j 0.9053, 0.3058 + j 1.2245, 0.0394 + j 0.3192, 0.6808 + j 1.7582.

4.5 Summary

173

The apparent power consumed by the impedances, the product of the current, its conjugate and the impedance divided by 2 are 3.9958 + j 0.3996, 1.3357, 0.0079 + j 0.0789, 0.1552, 1.7775. The total apparent power consumed by the circuit is the sum of those of all the impedances, S = Pav + j Qav = 7.2720 + j 0.4784 va. The power supplied by the sources is also the same. (−0.7195 − j 0.3207) + (7.9915 + j 0.7992) = 7.2720 + j 0.4784. The first component is the power supplied by the voltage source. Now, let us compute the power consumed by the resistors alone, which is the active average power Pav . The resistors are 1, 3, 1, 3, 1 and the corresponding currents are the same as the branch currents. The total power consumed by the resistors is 3.9958 + 1.3357 + 0.0079 + 0.1552 + 1.7775 = 7.2720 W and is equal to Pav found earlier. pf =



lagging.

4.4

Pav = 0.9978 |S|

Application

While electric power provides comfort in home, office, and industry, of course, we have to pay for it. Periodically, the reading of the kilowatthour meter is taken and we are billed for our use of power by the utility. A kilowatthour meter is used to measure the consumption of electric power. It is located between the power lines and the distribution panel of the building. The energy is the product of rate of power and the time period over which the power is used and its unit is wattseconds or joules. Since wattseconds is too small, watthour and kilowatthour are often used. One kilowatthour (kWh) is the energy dissipated by a 200 W bulb in 5 h. Typical wattage ratings of a laptop computer, washing machine, and a smoothing iron are, respectively, 70, 500, and 1000. For example, using a laptop for 7 h a day requires (70 × 7)/1000 = 0.49 kWh of energy.

4.5

Summary

• Electric power is the rate at which electric energy is transferred by an electric source to a sink that absorbs power. • Circuit element resistor dissipates energy. Capacitors and inductors store energy, respectively, in their electric and magnetic fields and, hence, are called storage elements. • The stored energy is returned to the source. Therefore, the average power consumed by the storage devices, in the sinusoidal steady state, is zero.

174

4 Steady-State Power

• The power in AC circuits has two components to take into account, as the sinusoid and the impedance are characterized, at a frequency, by two parameters, the magnitude and the phase. • The instantaneous energy stored in an inductor is 12 Li 2 (t). The instantaneous energy stored in a capacitor is 12 Cv 2 (t). • Energy measured in joules is the integral of power over an interval. A resistor always absorbs power. The average power absorbed by inductors and capacitors is zero. • The instantaneous power absorbed by the circuit is s(t) = 0.5V I cos(θ − φ) + 0.5V I cos(2ωt + θ + φ). • In the frequency-domain representation of circuits, the voltages and currents are, respectively, represented as V  θ and I  φ. Then, 0.5(V  θ )(I  (−φ)) = 0.5V I  (θ − φ) = 0.5V I (cos(θ − φ) + j sin(θ − φ)). The first component, which is the power actually consumed, is called the active power Pav . The power that is swapped back and forth from the source and the circuit is called the reactive or wattless power Qav . • Since the reactive power may be positive or negative, it is possible to change its value by adding suitable passive circuits, a process called power-factor correction. • The equivalent definitions of the power factor are pf =

R active power = cos(θ ) = . |apparent power| |Z|

Angle θ is the phase angle of the current with respect to the voltage. • For inductive loads, with a voltage source, a suitable parallel capacitor C has to be connected to improve the power factor. It is possible to improve the power factor for capacitive loads by connecting an inductor L in parallel.

Exercises 4.1 Consider the circuit, shown in Fig. 4.10, with the voltage applied being v(t) = sin(2t) V .

Fig. 4.10 A RL series circuit

sin(2t)V

The impedance of the circuit is composed of a resistor 2 connected in series with an inductor 1 H. Find the apparent, active, and reactive powers and the pf of the circuit. If the pf is to be 0.8

+ ∼

1H 2Ω

Exercises

175

cos(t)A

Fig. 4.11 A RC series circuit

∼

0.5F 1Ω

and 1, what is the value of the capacitor to be connected in parallel with the impedance in each case? * 4.2 Consider the circuit, shown in Fig. 4.11, with the current source i(t) = cos(t) A. The impedance of the circuit is composed of a resistor 1 connected in series with a capacitor 0.5 F. Find the apparent, active, and reactive powers and the pf of the circuit. If the pf is to be 0.8 and 1, what is the value of the inductor to be connected in series with the impedance in each case? 4.3 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.50. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Find the value of the capacitor for power-factor improvement, for pf = 0.8 lagging and pf = 1. * 4.4 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.53. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Find the value of the capacitor for power-factor improvement, for pf = 0.8 lagging. 4.5 Find the average active and reactive power consumed by the circuit, shown in Fig. 3.55. What is the power factor of the circuit? Verify that the power supplied is equal to the sum of the power consumed by the components. Find the value of the capacitor for power-factor improvement, for pf = 1.

5

Magnetically Coupled Circuits

5.1

Mutual Inductance

The circuit elements in circuits, we analyzed thus far, are conductively coupled. Conductors are used to connect elements. Circuit elements can also be connected through magnetic fields, called magnetically coupled. An inductor is basically a conducting coil. The induction of a voltage in another inductance by the current flowing in an inductance is mutual induction. A flywheel is a mechanical device that stores rotational energy and returns it to the system. Similarly, an inductor stores magnetic energy and returns it to the system. It does not dissipate energy like a resistor. While a resistor is required to limit the current in the circuit, an inductor smooths the flow of current. A time-varying current i(t) flowing through an inductor of value L11 henries induces a voltage, which tends to oppose the current increase, di(t) . v(t) = L11 dt The positive value L11 of the inductance is its self-inductance, or inductance because the voltage induced is due to the current flowing through itself, as we have seen earlier. A voltage v(t) = L12

dx(t) dt

is also induced in the inductor, which is affected by the magnetic field of another inductance due to current x(t). The mutually induced voltage may be positive or negative depending upon the relative directions of the induced voltages. The value L12 is called the mutual inductance between the two inductances and also measured in henries. A current in an inductance may induce voltages in a number of other inductances in its proximity, in addition to its self-induced voltage, and the open-circuit mutual voltages are designated as v1 (t) = L11

di1 (t) , dt

v2 (t) = L21

di1 (t) , dt

v3 (t) = L31

di1 (t) ,.... dt

In the determination of the mutual inductances, only a pair of coils is considered at a time, assuming that no other induced currents in the other inductances (by open-circuiting others). The study of mutual inductance is mandatory in analyzing circuits with often used components such as induction motors, transformers, induction heaters, etc. © Springer Nature Switzerland AG 2020 D. Sundararajan, Introductory Circuit Theory, https://doi.org/10.1007/978-3-030-31985-4_5

177

178

5 Magnetically Coupled Circuits

The currents and voltages induced in various inductances (for example with three inductances) are related as di1 (t) di2 (t) di3 (t) + L12 + L13 dt dt dt di1 (t) di2 (t) di3 (t) v2 (t) = L21 + L22 + L23 dt dt dt di1 (t) di2 (t) di3 (t) v3 (t) = L31 + L32 + L33 . dt dt dt

v1 (t) = L11

The self-inductances and the mutual inductances are, respectively, {L11 , L22 , L33 } and

{L12 , L13 , L21 , L23 , L31 , L32 }

with Lij = Lj i . In these expressions, i1 , i2 , i3 are the respective currents and v1 , v2 , v3 are the respective voltage drops. As Lij = Lj i , the mutual inductance between two coils is also denoted by M.

5.2

Stored Energy

Consider the case of two mutually coupled inductors with their coefficients being {L11 , L12 , L21 , L22 } with L12 = L21 . Then, equations relating their voltages and currents are di1 (t) di2 (t) + L12 dt dt di1 (t) di2 (t) v2 (t) = L21 + L22 . dt dt v1 (t) = L11

(5.1) (5.2)

Multiplying the two equations, respectively, by i1 (t) and by i2 (t) and adding, we get v1 (t)i1 (t) + v2 (t)i2 (t) = L11 i1 (t)

di1 (t) di2 (t) di1 (t) di2 (t) + L12 i1 (t) + L21 i2 (t) + L22 i2 (t) dt dt dt dt

Let n  1  1 2 2 L11 i1 (t) + L12 i1 (t)i2 (t) + L21 i1 (t)i2 (t) + L22 i2 (t) = T (t) = Lpq ip (t)iq (t) 2 2 p,q=1

where n is the number of coupled inductors. Then, with n = 2, we get dT (t) di1 (t) di2 (t) di1 (t) di2 (t) = L11 i1 (t) +L12 i1 (t) +L21 i2 (t) +L22 i2 (t) = v1 (t)i1 (t)+v2 (t)i2 (t) dt dt dt dt dt The instantaneous power absorbed by the two coils is equal to the derivative of the energy T stored in the associated magnetic fields.

5.2 Stored Energy

179

The equation for T (t) is in quadratic form. That is,  

  L11 L12 i1 (t) = L11 i12 (t) ± 2L12 i1 (t)i2 (t) + L22 i22 (t). 2T (t) = i1 (t) i2 (t) L21 L22 i2 (t) The energy stored in the magnetic fields must be positive regardless of the signs of the currents. The sign is positive for the mutual inductance term, if both the currents leave or enter the dotted terminals of the coils (as shown in Fig. 5.1). Otherwise, the energy contribution is negative. A quadratic form is positive definite if all the principle minors in the top-left corner of its corresponding matrix are all zero. That is,   L L  |L11 > 0,  11 12  > 0. L21 L22 Then, L11 > 0,

L22 > 0,

and

L11 L22 − L212 > 0 or

L212